PRACTICAL MATHEMATICS 

HOBBS-WAITE-SCHROETER 




CbpghtN! 



COPYRIGHT DEPOSIT. 



PRACTICAL MATHEMATICS 



AX ELEMENTARY TREATISE COVERING THE FUNDAMENTAL 
PROCESSES OF ARITHMETIC, ALGEBRA, AND GEOM- 
ETRY, WITH A PRACTICAL PRESENTATION 
OF LOGARITHMS AND CURVE 
PLOTTING 



By 
GLENN M. HOBBS, Ph. D. 

SECRETARY AND EDUCATIONAL DIRECTOR 
AMERICAN SCHOOL OF CORRESPONDENCE 



EDWARD B. \yAITE 

DEAN AND HEAD, CONSULTING DEPARTMENT 

JOHN P. SCHROETER 

CONSULTING ENGINEER, DEPARTMENT OF 
ELECTRICAL ENGINEERING 



ct^vTD 



AMERICAN TECHNICAL SOCIETY 

CHICAGO 

1915 






COPYRIGHT, 1915, BT 

AMERICAN TECHNICAL SOCIETY 



COPYRIGHTED IN GREAT BRITAIN 
ALL RIGHTS RESERVED 



LC Control Number 



JAN 19 1915 
©CLA393348 


1 

tin 


mil 

p96 02^ 


III 

'227 



CONTENTS 

v) — 



FUNDAMENTAL PROCESSES 

PAGE 

Definitions 2 

Mathematics 2 

Quantity 2 

Unit 2 

Concrete number 2 

Abstract number 2 

Like quantities 2 

Mathematical signs 2 

Addition 2 

Subtraction 2 

Multiplication 2 

Division 3 

Equality 3 

Parenthesis 3 

Vinculum 3 

Notation 3 

Roman 3 

Arabic 4 

Numeration 5 

The orders 6 

The periods 6 

Addition 7 

Rules 8 

Proof 9 



Subtraction 

Minuend 

Subtrahend 

Difference or remainder. 
Rules 



2 CONTENTS 

PA.GB 

Multiplication 14 

Multiplicand 14 

Multiplier 14 

Product 14 

Coefficient .' 15 

Rules 15 

Short methods 17 

Checking, method of 18 

Division 20 

Dividend 20 

Divisor 20 

Quotient 20 

Long division 20 

Short division 21 

Rules 21 

Proof 22 

Short methods 22 

Factoring 24 

Prime number 24 

Prime factor 24 

Factoring by observation -25 

Cancellation ' 26 

Methods 26 

Rules 28 

FRACTIONS 

Definitions 31 

Denominator 31 

Numerator 31 

Fractional unit 32 

Value 32 

Proper fraction 32 

Improper fraction 32 

Reduction of fractions 33 

Entire quantity 34 

Mixed quantity 34 

Improper fraction to mixed quantity, reduction of 34 

Least common denominator 35 

Multiple 35 



CONTENTS 3 

PAGE 

Addition of fractions 39 

Mixed quantities 41 

Subtraction of fractions 42 

Mixed quantities 42 

Multiplication of fractions 44 

Division of fractions 46 

Decimals 49 

Pure decimal 50 

Mixed decimal 50 

Addition 52 

Subtraction 52 

Multiplication 53 

Division 5o 

Per cent 57 

Base 59 

Rate 59 

Percentage 59 

Rules 59 

DENOMINATE NUMBERS 

Measures 63 

Unit of measure 63 

English system 63 

Metric system 63 

Measures of extension 64 

Linear measure 64 

Square measure 65 

Cubic measure 66 

Measures of capacity 67 

Liquid measure 67 

Dry measure. .' 68 

Measures of weight 68 

Troy 68 

Apothecaries 68 

Avoirdupois 68 

Metric 68 

Measures of time 69 

Mean solar day 69 

Standard times 69 



4 CONTENTS 

PAGE 

Miscellaneous measures 70 

Reduction of denominate numbers 70 

Rules 70 

Operations with denominate numbers 72 

Addition and subtraction 7.2 

Multiplication and division 73 

ADVANCED PROCESSES 

Powers and roots 74 

Power, definition of 74 

Root, definition of 76 

Radical sign 76 

Square root 77 

Square root of fractions 79 

Ratio and proportion 80 

Ratio, definition of 80 

Antecedent 80 

Consequent 80 

Direct ratio 81 

Inverse ratio 81 

Proportion, definition of 82 

Means 82 

Extremes 82 

Rules 83 

Negative quantities 86 

Parenthesis 88 

Rules 88 

Equations 89 

Transposition 90 

Equations containing fractions 92 

Rules 93 

General application to engineering problems. 95 

PRACTICAL GEOMETRY . 

Lines 99 

Straight 99 

Curved 99 

Broken 99 

Plane 99 

Parallel ' 99 



CONTENTS 5 

PAGE 

Angles 99 

Sides 99 

Vertex 99 

Right 99 

Acute 100 

Obtuse 100 

Measurement 100 

Polygons 102 

Triangles 102 

Quadrilaterals 109 

Areas of other polygons 110 

Circles Ill 

Areas of circles 112 

Areas of sectors and segments 113 

Solids 115 

Prisms 115 

Cylinders 117 

Pyramids 118 

Cones 120 

Spheres 122 

LOGARITHMS 

Method of Calculation 124 

Characteristic and mantissa 124 

Finding logarithm of a number 126 

Finding number from its logarithm 127 

Applying Logarithm Tables 128 

Multiplication by logarithms 128 

Division by logarithms 129 

Powers 130 

Roots 130 

Rules 132 

CURVE PLOTTING 

Plotting methods 133 

Co-ordinate paper 133 

Relation of co-ordinates to curve 134 

Plotting the curve 136 



6 CONTENTS 



PAGE 



Meaning of curves 139 

Study curves 139 

Horsepower curve 139 

Illustrative curves 140 

Curves for squares and square roots 140 

Curve for calculating length of belt 142 

Gross income curve for steam railroad 143 

Contract price for sewer pipe 144 

Mean precipitation curve 145 

Speed curve for steamboat 146 

Curve showing fire loss per capita 147 

Sunday current load curve 147 

Locomotive repair curve 147 

Water power and load curves 148 

Comparison curves for incandescent lamps 150 

Curve of wheat prices 152 

Speed curves for wheels or pulleys 152 

APPENDIX 

Tables 156 

Linear measure 156 

Square measure 156 

Cubic measure % 156 

Surveyors' linear measure 156 

Surveyors' square measure 157 

Dry measure 157 

Liquid measure 157 

Apothecaries' fluid measure -:. . 157 

Troy weight 158 

Avoirdupois weight 158 

Long ton weight 158 

Apothecaries' weight 158 

Metric system 159 

Measures of time 162 

Measures of money 162 

Miscellaneous measures 163 

Slide rules 164 

Logarithm tables 170 



INTRODUCTION 



HPHE methocTof teaching mathematics has been so well organized 
A in connection with the public schools of the country that very- 
few changes are made from year to year. However, for those 
who wish to use mathematics only as a means to an end, without 
thought of the cultural or training value of the subject, much 
material can without .detriment be eliminated from the average 
school arithmetic. It was, therefore, thought wise to give in this 
treatise only the essentials of the various processes, with special 
emphasis on the practical side of the subject. 

<I It will be noted that the plan of using algebraic symbols in 
connection with the treatment of even the elementary processes 
has been adopted, the feeling of the authors being that the mystery 
of Algebra is the hardest part of the subject and that by a sort of 
homeopathic acquaintance with the symbols as substitutes for 
figures, a greater confidence would be established in the student's 
mind when the actual algebraic problems were taken up. 

<I Another innovation is the introduction of a simple presenta- 
tion of logarithms as a useful method of calculation, particularly as 
a handy method of handling powers and roots. No attempt has 
been made to give the theory of logarithms, but a great many 
examples illustrating the process are given and a simple four-place 
table is included in the book. 

<I Finally, a simple discussion of curve plotting as applied to 
dynamo and efficiency curves and other practical application of 
the graphical method of representing mathematical results is 
given. The practical justification of such a step is easy to estab- 
lish as workmen, engineers, and even the reading public, are often 
called upon to interpret or at least to understand curves of this 
nature. 



PRACTICAL MATHEMATICS 

PART I 



INTRODUCTION 



Xo one who is at all acquainted with the demands of the En- 
gineering profession will deny the need of a good foundation in 
elementary mathematics any more than he will deny the need of a 
solid underpinning on which to rest the walls of a big business block. 

The simplest problems of the contractor and workman,- such as 
the number of feet of lumber required for a house, the number of 
cubic yards of excavation for a ditch or cellar, the proper under- 
standing of plans and specifications, and the laying off of measure- 
ments according to these plans, all require a knowledge of this im- 
portant subject. The size of a concrete retaining wall, the dimensions 
of a girder for a steel structure, the amount of iron in the field of a 
dynamo, or the capacity of the cylinders of an engine, is certainly 
not left to the arbitrary judgment of a foreman but is carefully worked 
out by mathematics and by a knowledge of the properties of the 
materials used. 

Mathematics might be likened to a kit of tools which the work- 
man carries; the master workman carries more than the apprentice 
and the more tools each man has in his kit and knows how to use, 
the more things he can do and the greater is his earning power. 
Each mathematical process is a tool to be used as the occasion de- 
mands. Some of them are used in every problem which comes up, 
others less frequently, but the more advanced the work the greater 
the number of tools required. 

It is with this keen demand in mind, therefore, that we are 
requiring of each student at the outset of his course this work or its 
equivalent in Practical [Mathematics. We want him to fill his kit 
with enough tools to meet the steady demands of the work ahead of 
him, and we feel sure that, once provided with this equipment, his 
progress will be assured. 



2 PRACTICAL MATHEMATICS 

In the preparation of this work the authors have intentionally 
lost sight of the material usually found in the school books on this 
subject, and have kept in mind only the particular parts which are 
of special importance to the engineering student. Not only the 
topics discussed but all of the problems have been made exception- 
ally practical, and the aim has been at all times to give the student 
the satisfaction of knowing that whatever he is learning will be of 
use in his work and will also count for his advancement. 

DEFINITIONS AND MATHEMATICAL SIGNS 

1 . Definitions. Mathematics is the science which treats of quan- 
tity ; its fundamental branches are Arithmetic, Algebra, and Geometry. 

Quantity is anything which can be increased, diminished, or 
measured; as numbers, lines, space, motion, time, volume, and 
weight. 

A unit is a single thing, or one. 

A number, or quantity, is a unit or a collection of units and is 
either concrete or abstract. 

A concrete number, or quantity, is one whose units refer to par- 
ticular things, as, for example, 5 rivets, 7 bolts. 

An abstract number, or quantity, does not refer to any particular 
thing. For example, 5, 23, etc., used without designating any 
particular objects, are abstract numbers. 

Like quantities refer to the same thing, as 7 saws, 2 saws; unlike 
quantities refer to different things, as 2 trunks, 4 tables, 3 chairs. 

2. Mathematical Signs. For the sake of brevity, signs are 
used in mathematics to indicate processes. Those signs most" used 
in Arithmetic are +, — , X, -f-, —> ()> and 

The sign + is read "plus" and is the sign of addition. It shows 
that the quantities between which it is placed are to be added 
together. If 2 and 2 are to be added, it is expressed, thus: 2+2 are 
four. 

The sign — is read "minus" and is the sign of subtraction. It 
means that the quantity which follows this sign is to be subtracted 
or taken away from the quantity which precedes it, thus: 6 — 4 are 2. 

The sign X is read "times" and is the sign of multiplication. 
It means that the quantity which precedes this sign is to be multiplied 
by the quantity which follows it, thus: 2X5 are 10. 



PRACTICAL MATHEMATICS 3 

The sign -5- is read "divided by" and is the sign of division. 
It means that the quantity which precedes this sign is to be divided 
by the -quantity which follows it, thus: 4-r2 arc 2. 

The sign = is read "equals" or "is equal to" and is the sign of 
equality. It means that the expressions between which it is placed 
are identical in value, thus: 4 + 3 = 10 — 3. This sign is very 
often misused. Great care should be taken at all times to make sure 
that the quantities connected by it arc equal. For example, it would 
be absurd to say that 5 + 9 = 14 -~ 2 = 7, because 5 + 9 does not 
equal 7. 

The parenthesis ( ) and vinculum " ~ are used to show 

that two or more quantities are to be treated as one; or in other 
words, that the operations indicated within the parenthesis or 
under the vinculum are to be carried out first, thus: — (20+5) is 
equal to —25, not —20+5; similarly, —2 + 3 is equal to —5. 

NOTATION 

3. Xotation is the art of writing numbers in words, in figures, 
and in letters. 

There are two methods of notation in common use; the Roman 
and the Arabic. 

4. Roman Notation. In the Roman notation, 7 capital letters 
are used, as follows: 

I is used to express one. 



V" 


it 


t a 


five. 


X" 


tt 


i it 


ten. 


L" 


tt 


it tt 


fifty. 


C" 


tt 


i a 


one hundred. 


D " 


tt 


a a 


five hundred. 


M" 


a i 


t it 


one thousand. 



All other numbers are expressed by repetitions or by combina- 
tions of these seven letters according to the following rules: 

By repeating a Utter the value denoted by the letter is doubled; 
thus: XX means twenty; CC means two hundred. 

By placing a lefirr denoting a less value before a letter denoting 
a greater, their difference of value is represented; thus: IV denotes 
four or one less than five; XL denotes forty or ten less than fifty. 



PRACTICAL MATHEMATICS 



By placing a letter denoting a less value after a letter denoting a 
greater valve, their sum is represented; thus: VII denotes seven or 
two more than five. XV denotes fifteen or five more tha ! n ten.' 

A line " " placed over a letter increases the value denoted by 
the letter a thousand times; thus: X means ten thousand. IV means 
four thousand. 

The use of the Roman notation is now confined mainly to the 
writing of dates, and the numbering of chapters in books, and the 
hours on the dials of clocks. 

Table I shows some of the combinations of the 7 letters used. 

TABLE I 
Roman Notation 



I 


one 


LXXX 


eighty 


II 


two 


XC 


. ninety 


Ill 


three 


C 


one hundred 


IV 


four 


CC 


two hundred 


V 


five 


CCC 


three hundred 


VI 


six 


D 


. five hundred 


VII 


seven 


DC 


six hundred 


VIII 


. eight 


DCC 


seven hundred 


IX 


nine 


CM 


nine hundred 


X 


ten 


M 


one thousand 


XX 


twenty 


MD. 


fifteen hundred 


XXX 


thirty 


MM 


two thousand 


XL 


forty 


X 


ten thousand 


L 


fifty 


M 


one million 


LX 


sixty 


MCMX 


1910 


LXX 


seventy 







5. Arabic Notation. The Arabic notation employs ten char- 
acters or figures in expressing numbers. They are 

1, 2, 3, 4, 5, 6, 7, 8, 9, 

one two three four five six seven eight nine cipher 

The first nine are sometimes called digits; the cipher is also called 
naught or zero because it expresses nothing or the absence of a number. 

The digits (1, 2, 3, 4, 5, 6, 7, 8, 9) have been termed significant 
figures because each has of itself a definite value, always represent- 
ing so many units or ones as its name indicates. However, the value 
of the units represented by a figure depends upon the particular 
position which that figure occupies with regard to other figures. This 
position is called its place or order. 



hundrei 

tens 

units 




4 = 


4 units 


4 = 


4 tens 


4 = 


4 hundreds 


4 4 4 = 


: total 



PRACTICAL MATHEMATICS 5 

For example, if three figures are written together to represent a 
number, as 444, each of these figures, without regard to its place, 
expresses four units, but when considered as 
part of the number these fours differ in value. 
The 4 in the first place to the right represents 
4 units; the 4 in the second place, represents 
4 tens or 4 units each ten times the size or 
value of a unit of the first place; and the 4 in 
the third place, represents 4 hundreds, or 4 
units each one hundred times the size or value 
of a unit of the first place. It is readily seen that the value of any 
figure is increased ten-fold by removing it one place to the left. 

The cipher becomes significant when 
connected with other figures by filling a place 
that otherwise would be vacant, as in 10 (ten) 
it gives a ten-fold value to the 1. In 130 
(one hundred thirty) it gives a ten-fold value 
to the 13. A cipher between two or more 
figures produces the same effect. In 405 the 
cipher which fills the intervening place be- 
tween 4 and 5 causes the 4 to represent four hundreds, not four tens. 

The following principles should be firmly fixed in mind: 

.4// numbers are expressed by the nine digits and zero. 

Zero has no value; it is used to fill vacant places only. 

A figure has different values according to the place it occupies. 

The base of the system of notation is ten; ten units of any order 
making one unit of the next higher order. 

NUMERATION 

6. Numeration is the art of reading numbers when expressed 
by letters or figures. 

This is accomplished by first enumerating the orders from 
right to left, as shown in Table II, and then reading these orders 
in the reverse direction in groups of three, called periods. The 
first three orders, Units, Tens, Hundreds, constitute the first or 
unit period. The second three orders form the second or thousand 
period; the third three orders, the third or million period; and so on.* 

* Commas are used to separate the different periods of figures. In reading numbers 
never use "and" to connect the periods. 



hundreds 

tens 

units 




5 = 


5 units 


= 


: tens 


4 = 


: 4 hundreds 


4 5 = 


total 



6 PRACTICAL MATHEMATICS 

The system of periods is shown clearly in Table III. Such a number 
as 534 consists of one period and is read — five hundred thirty-four. 
The number shown below the orders in Table II is read — two million, 
seven hundred fifteen thousand, six hundred thirty-nine. 

TABLE II 
The Orders 



7th 


6th 

co 

a 

a 

CO 

O 

H 

CM 


5th 

CO 

C 
c3 

CO 

o 


4th 


3rd 


2nd 


1st 




co 
S3 


o 

CO 

(V 




o 


•3 

a3 


•3- 

CD 








O 


-a 


co 
S3 

0) 


3 

O 

J3 


3 


CO 

a 

CD 


•3 




§ 


W 


h 


H 


w 


Eh 


P 




2 


7 


1 


5 


6 


3 


9 





The number in Table III, which is 987654321987654, is divided 
into the periods 987, 654, 321, 987, 654, and is read — nine hundred 
eighty-seven trillion, six hundred fifty-four billion, three hundred 
twenty-one million, nine hundred eighty-seven thousand, six hun- 
dred fifty-four. 

TABLE III 

The Periods 



*J 




53 °a 



T3 T3 

£ £ 

Jga Jga pga sga sga 

Whp Whp Khp Whp KhP 

987 654 321 987 654 

5th 4th 3rd 2nd 1st 

Period Period Period Period Period 



PRACTICAL MATHEMATICS 
PROBLEMS FOR PRACTICE 



Write 


in words: 




1. 


18,765,972. 




2. 


834,769,780. 




3. 


3,576,879,42 


1. 


4. 


10,805,056. 


Ans. 


"Write in figures: ■ 





Ten million, eight hundred five thou- 
sand, fifty-six. 



5. Seventy-eight million, forty-one thousand seven. 

6. One thousand three. Ans. 1,003. 

7. Five hundred six thousand. 

S. Ninety million, two thousand, three hundred twenty-seven. 

9. Three hundred five thousand seventy-nine. 

10. Eight hundred sixty-four million, four thousand, twenty. 

ADDITION 

7. Adding is the process of finding a number which is equal to 
the combined values of two or more given numbers. The result thus 
obtained is called the sum. Hence, it may be said that the sum of 
two or more numbers is a number containing as many units as all 
the numbers taken together. Thus the sum of 5 rivets and 7 rivets 
is 12 rivets, since 12 contains as many units as 5 and 7 together. 

Letters of the alphabet are also used to represent quantities. 
A letter may stand for any number, but the same letter must have 
one value throughout a given problem, although it may have differ- 
ent values in different problems. For example, a may equal 3 in 
one problem, and 7 or 13 in another; b may equal 2 in one problem, 
and 5 in another; c may equal 2, 3, 9 or 25; and so on with other 
letters. The sum of a, b, and c will be equal to a+b-\-c, just as the 
sum of 2, 5, and 7 will be equal to 2 + 5 + 7. Now, if a = 2, 6 = 5, 
and c = 7, the sum of a, b, and c is found by simply adding together 
2, 5, and 7, obtaining 14; on the other hand, if the values of a, 6, and 
c are not known, their sum can only be indicated. If a and b rep- 
resent the horse-powers of two different engines or of two weights 
lying in a scale-pan, then a+b equals the total horse-power or the 
total weight. 



8 PRACTICAL MATHEMATICS 

8. Since a number is a collection of units of the same kind, 
two or more numbers may be united into one sum only when they 
are like quantities; thus: 2 bolts + 4 bolts = 6 bolts, but 4 rivets 
and 3 bolts cannot be united into one quantity, either of bolts or of 
rivets. On the other hand, if quantities of different kinds are 
represented by letters, they may be added together in order to help 
solve the problem. For example, if a represents a car ) and b an 
engine, their sum may be indicated (a + b). It must be remembered, 
however, that the car and the engine have not actually been added 
together, because they represent unlike quantities. The use of 
letters makes the solution of certain kinds of problems much simpler, 
and later will be taken up more fully. 

Rules for Addition. Write the numbers under each other, placing 
them so that units are under units, tens under tens, hundreds under 
hundreds, and so on. 

Add up the column of units; and put the right-hand figure of this 
sum under the unit column, carrying the remaining figure or figures 
to the column of tens; add up the tens column, including the carried 
figures, put down the right-hand figure and carry as before. Continue 
in this way until the last column is reached, putting down the total 
of the last column to give the final sum. 

Examples. 1. Find the sum of 567; 141; and 93. 

Solution. Write these numbers under one another, so that 
the units of each shall be in the same vertical column. Then add 
up as follows: 3 and 1 are 4, and 7 are 11. Place the right-hand ^®' 

figure 1 under the units column and carry 1 ten to the next column. 141 

Adding the tens column, 1 (carried) + 9 are 10 and 4 are 14 and 93 

6 are 20. Put down the right-hand figure, which is zero, and carry 801 

2 to the next column; then 2 (carried) and 1 are 3 and 5 are 8. 
Putting down the 8, the total sum is found to be 801. 

2. Find the sum of 6,321; 2,576; 9,702; and 257. 

Solution. Write these numbers one under the other as be- 
fore. Add up the unit column as follows: 7 and 2 are 9, and 6 are aooi 
15, and 1 are 16. Put down the 6 and carry the 1; then adding the OoZl 
tens column, 1 (carried) and 5 are 6, and 7 are 13, and 2 are 15. ^57o 
Put down the 5 and carry the 1; then adding the hundreds column, 9/02 
1 (carried) and 2 are 3, and 7 are 10, and 5 are 15, and 3 are 18; put 257 
down the 8 and carry the 1. Then adding the thousands column, 18856 
1 (carried) and 9 are 10, and 2 are 12, and 6 are 18. Putting down 
the whole amount, 18, gives the total sum of the numbers as 18,856. 



PRACTICAL MATHEMATICS 9 

3. Add the following quantities: a+2b+ c, 36 + 2c, 6a -f- 

6 + c, and 106. 

Solution. Write the quantities in columns, keep- 
ing the like terms under each other; that is, keep the «'s in 

the first vertical column, the fr's in the second, the c's in a "r ~D H~ C 

the third. The sum of the a's is a 4- 6a or 7(7.* 26 + 36 36 + 2c 

4-6 4- 10!) are 166. c - 2e - c are 4c. Connecting the 6a + 6 + C 

sums by the c<ign of addition the total is found to be 7<t 4- 106 

166 4- 4c. When the number showing how many times ~7a -j- 166 + 4c 
the letter is taken is omitted, the letter is to be taken 
once. Thus: a = la, 6 = 16, c = lc. 

9. Proof. To prove that a sum is correct, begin at the top and 
add the columns downward in the same manner as they were added 
upward; if the two sums agree, the work is presumably correct, for 
adding downward inverts the order of the figures, and therefore 
any error made in the first addition would probably be detected in 
the second. 

PROBLEMS FOR PRACTICE 

Find the sum of 

1. 56 + 49 + 17 + 36 + 21. Ans. 179 

2. 42 - 46 + 43 + 58 + 91. 

3. 467 + 536 + 84 + 705. Ans. 1,792 

4. 2,008,+ 1,400 + 706 + 300 + 77. 

5. 8,950 + 15,765 + 7,732. Ans. 32,447 

6. 26,661 + 8,735 + 6,877 + 33,413. 

7. 8,792 + 980 + 5,607 + 89. 

8. 346 + 4,682 + 64 + 798 + 21. 

9. 26 + 425,902 + 3,006 + 490 + 36,221. Ans. 465,645 

10. 3a + 116 4- a + la + 56. Ans. 11a + 166 

11. c + 2a . + 5c + 4a + 3c. 

12. 6a + 6 + bd + 2d. Ans. 6a + 6 + 7d 
L3. 4a + 26 + 3a + 7c + 3c. 

10. It has now been shown how to add numbers correctly 
when the process is indicated. It often happens, however, that prob- 
lems will be met in which only the statement of the relations between 
quantities are given. The following illustrative example will show 
the method of solving problems of this nature. 

* 7 is called the coefficient of a and indicates that a is to be taken seven times. See 
J 14, p. 15. 



10 PRACTICAL MATHEMATICS 

Example. An electric light plant, capable of furnishing current 
for 200 16-candle-power lamps cost as follows: 

16 Horse-power engine 350 dollars 

Dynamo 275 " 

Driving belt 50 

Installation 35 

What was the total cost? 

Solution. In an example of this kina the problem should 350 

be read several times until it is thoroughly understood. If the 275 

numbers to be used in computation are fixed in mind the student en 

will not be misled by the wording. For example, it can readily be oc 

seen that 200 will not be used in the computation. Find the total 

cost of the equipment as shown in the margin. ' ■*•" 

PROBLEMS FOR PRACTICE 

1. A marine engine during a 3 hours' run makes 9,187 revolu- 
tions the first hour, 9,062 the second, and 9,233 the third. How 
many does it make in the 3 hours? 

2. Coal is fed to a furnace as follows: Monday, 376 pounds; 
Tuesday, 307 pounds; Wednesday, 438 pounds; Thursday, 425 
pounds; Friday, 399 pounds; Saturday, 301 pounds. Find the total 
for the week. 

3. The items for lumber called for in a contract w 7 ere — frame, 
3,896 feet; flooring, 6,796 feet; finish, 2,739 feet. How many feet 
were used? 

4. A surveying party works six weeks. The first week they 
survey 151 miles; the second week, 111 miles; the third week, 162 
miles; the fourth week, 159 miles; the fifth week, 96 miles; the sixth 
week, 48 miles. How many miles did they survey? 

5. There are five water wheels installed in a water power plant. 
The power furnished by the first wheel is 2,225 horse-power, and 
the others furnish, 3,150, 4,275, 5,650, and 8,275 horse-power. 
What is the total capacity of the five wheels? 

6. The weekly capacity of 4 lathes is as follows : 2,500 castings, 
4,175 nuts, 3,420 brass boxings, and 2,185 finished trimmings. How 
many pieces do the four lathes turn out per week? 

7. When purchasing an 85 dollar indicator, the following extras 
were bought: 






PRACTICAL MATHEMATICS 11 

1 Spring 5 dollars 

1 Elbow 3' " 

Pantograph 10 

Speed Counter 2 " 

Planiineter 20 

What was the total cost of the outfit including the indicator? 

v If a 10-inch belt will transmit 17 horse-power at a speed of 
1 ,800 feet per minute, and a 16-inch belt will transmit 36 horse-power 
at the same speed, how much power will be transmitted by the two 
belts? 

SUBTRACTION 

11. Subtraction is the process of finding the difference between 
two quantities; this difference when added to the smaller will give a 
result equal to the greater. 

For example, the difference between 16 and 7 is 9, since 7 added 
to 9 makes 16. The greater of the two quantities whose difference 
is to be found, is called the minuend; the smaller is called the sub- 
trahend. The quantity left after taking the subtrahend from the 
minuend is called the difference or remainder. 

Only like quantities and units of the same order can be sub- 
tracted, and the remainder is always like the minuend and sub- 
trahend. 

It can readily be seen that subtraction is the reverse of addition, 
and this fact is made use of to prove subtraction as shown in the 
following: 

Example. Subtract 114 from 237 and prove the result. 

Solution-. Beginning with the units column, 4 p f 

(units) are subtracted from 7 (units) leaving 3 (units), ^07 1 14 

which is set down directly under the column in units 1 . 

place. Proceeding to the next column 1 (ten) is sub- 

tracted from 3 (tens) leaving '1 (tens), which is set down 123 237 

in tena place. Proceeding as before, the final remainder 
is found to be 123. 

12. Rules for Subtraction. Write the subtrahend under the 

mi/iur, id 80 thai units of the stimr order will I"' in the same column. 

Begin with the units of the lowest order to subtract, and proceed 
tottie highest, writing each remainder under the line in its proper place. 



12 



PRACTICAL MATHEMATICS 



If any digit of the minuend is less than the corresponding digit 
of the subtrahend, add ten to it and then subtract; but consider that the 
next digit of the minuend has been diminished by one. 

Examples. 1. From 6,784 subtract 3,776. 



Solution. In order to subtract 6 in the units 
place in the subtrahend from 4 in the units place in the 
minuend, the 4 must first be increased by 10 giving 14; 
then 6 subtracted from 14 leaves 8 in the units place of 
the remainder. But the 10 added to the 4, has been ob- 
tained by diminishing the 8 in the tens place of the 
minuend by 1, leaving 7. The 7 in the tens place of the 
subtrahend subtracted from this leaves zero in the tens 
place of the remainder. The 7 in the hundreds place of 
the subtrahend subtracted from the 7 in the hundreds 
place of the minuend leaves zero in the hundreds place 
of the remainder, and the 3 in the thousands place 
of the subtrahend subtracted from the 6 in the 
thousands place of the minuend leaves 3 in the thou- 
sands place of the remainder, giving 3,008. 



6784 minuend 
3776 subtrahend 
3008 remainder 



2. From 1,000 subtract 621. 

Solution. The in the units place of the minu- 
end must first be increased by 10; then 1 subtracted 
from 10 leaves 9 in the units place of the remainder. In 
adding 10 to the 0, 1 has been taken from the tens place, 
but as it was itself 0, it had to borrow from the next place 
and continue borrowing until a numerical place was 
reached. Proceeding in this way the total remainder 
379 is obtained. 



1000 minuend 
621 subtrahend 
379 remainder 



13. Letters may be used in subtraction just as in addition. 
For instance, if a quantity represented by b is to be subtracted from 
a quantity represented by a, the difference will be represented by 
the quantity (a — b); or if c is to be subtracted from b, the difference 
will be represented by the quantity (b — c). If the quantity consists 
of a letter or letters with a coefficient, as Sab, and another term con- 
taining the same letters, as Sab, is to be subtracted from it, the sub- 
traction may be performed in the usual way. It must be remem- 
bered, however, that Sab means that ab is taken eight times, while 
Sab means that ab is to be taken three times. Therefore, if Sab 
is taken from Sab, bob remains. In other words, Sab — Sab equals 
5ab. 






PRACTICAL MATHEMATICS 13 

Examples. 1. From lab + 96 + 4c subtract 3a6 + b + 2c. 

Solution-. Arrange the tonus as in addition 
of letters, subtracting the coefficients only and bring- 
ing down the letter or loiters after each result. Tims: ' (, b ~r 9o + -4c 
7o6-3o6=4a6; 96-6=96-16=86; 4c-2c=2c. The 3a/> + 6 + 2c 
separate results of the subtraction may now be con- 4 a fr _|_ $b _l. 2c 
nected by the plus sign, giving as the final answer the 
quantity 4a6+86+2c 

2. From 6a + 2d subtract 5a + 2d. 

Solution. Subtracting we have 6a— 5a= la = a; ,, , 9 , 

2d - 2d = Od = 0. When the difference between the l _ )a ~J~ ~^ 

coefficients is 0, the term is 0. Therefore, the second oa -+- ~d 

term of the remainder disappears, giving as the final a 

result the quantity a. 



PROBLEMS FOR PRACTICE 

1. From 7,282 subtract 4,815 Ans. 2,467 

2. From 64,037 subtract 5,908. 

3. From 6,231 subtract 3,084. Ans. 3,147 

4. From 1,740,932 subtract 807,605. 

5. From 71,287 subtract 40,089. Ans. 31,198 

6. From 1,000,000 subtract 999,999. 

7. From the sum of 2,465,321 and 975,803 

subtract 739,034. Ans. 2,702,090 

8. From 96 + 7c subtract 46 + c. 

9. From 12a + 156 + 6c subtract 8a + 146 + 6c. 

10. From 5a + 26 subtract 3a + b. Ans. 2a + b 

11. From a + bb + 2d subtract a + 46 + 2d. Ans. b 

12. From a tank containing 935 gallons of water, 648 gallons 
were drawn off. Then 247 gallons ran in. How many gallons were 
then in the tank? (Suggestion: Subtract 648 from 935 and add 247.) 

L3. A man purchased 8,983 bricks, but used only 5,363. How 
many had he left ? 

14. A coal shed contains 8,579 tons. 3,243 tons are taken from 
it. It then receives 1,112 tons more. After that 1,602 tons are 
taken out of it. How many tons remain? 

15. An electric power plant can generate 2,000 horse-power. 
Of this, 1,910 horse-power is used. The manager then agrees to 



14 PRACTICAL MATHEMATICS 

furnish another firm with 784 horse-power. How much more power 
will he need? (Suggestion: Add 784 to 1,910 and subtract 2,000.) 

Ans. 694 horse-power. 

16. An engine develops 147 horse-power. 16 horse-power is 
used in running the engine itself. How much power is available for 
running machinery? 

17. 1,200 gallons are pumped from a tank. Of this, 22 gallons 
are lost in leakage, etc. How much is discharged by the pump? 

18. A 75 horse-power boiler evaporates 2,140 pounds of water 
into steam per hour. One engine uses 1,310 pounds, another uses 
417 pounds, and a pump requires the remainder. How much steam 
is used by the pump? (Suggestion: Add 417 to 1,310 and subtract 
from 2,140.) Ans. 413 pounds. 

19. An engine makes 54,000 revolutions in a day of 12 hours. 
A motor makes 720,000 revolutions in the same time. By how many 
revolutions per day does the speed of the motor exceed that of the 
engine? 

20. It takes 3,880 pounds of steam per hour to run a certain 1C0 
horse-power engine. If it takes 1,940 pounds to run a 60 horse- 
power engine and 2,140 pounds to run a 72 horse-power engine, 
does the largest engine require as much steam as the two small ones ? 
(Suggestion: Add 1,940 to 2,140 and compare with 3,880.) 

MULTIPLICATION 

14. Multiplication is a short method of adding a quantity to 
itself a certain number of times; or, it is the process of taking one 
quantity as many times as there are units in another. 

It is known that 2 + 2 + 2 + 2 + 2. = 10; but this same process 
may be expressed more briefly by the aid of multiplication, thus: 
5 X 2 = 10. The 5 shows how many twos are used in adding. 
This last expression is read, "five times two equals ten." 

In multiplication three terms are employed — the multiplicand, 
the multiplier, and the product. 

The multiplicand is the quantity to be multiplied or taken. 

The multiplier denotes the number of times the multiplicand is 
to be taken. 

The product is the result or quantity obtained by the multiplication. 



PRACTICAL MATHEMATICS 



15 



Another term, called the coefficient^ is used to indicate the 
numerical part of a quantity which consists of numbers and letters 
multiplied together. Thus in the expression 3a6, ^ is the coefficient 
of ah, and denotes that ab is taken 3 times. Sometimes it is con- 
venient to consider any part of the product as the coefficient of the 
remaining part. Thus in 3a&, 3a might be considered the coefficient 
of b t or 36 the coefficient of a. A coefficient is, therefore, called 
numerical or literal according as it is a number or one or more letters. 
When no numerical coefficient is expressed, 1 is always understood, 

TABLE IV 
Multiplication Table 



1 


2 


3 

6 


4 


5 


6 


7 


8 


9 


10 


11 


12 


13 


14 


15 


16 


17 


18 


19 


20 


21 


22 


23 

46 


24 

4S 


25 

50 


2 


4 


8 


10 


12 


14 


16 


18 


20 


22 


24 


26 


28 


30 


32 


34 


36 


38 


40 


42 


44 


3 


6 


9 


12 


15 


IS 


21 


24 


27 


30 


33 


36 


39 


42 


45 


4S 


51 


54 


57 


60 


63 


66 


69 


72 


75 


4 


S 


12 


16 


20 


24 


2S 


32 


36 


40 


44 


48 


52 


56 


60 


64 


68 


72 


76 


80 


84 


8S 


92 


96 


100 


5 


10 


15 


20 


25 


30 


35 


40 


45 


50 


55 


60 


65 


70 


75 


80 


85 


90 


95 


100 


105 


110 


115 


120 


125 


6 


12 


18 


24 


30 


36 


42 


48 


54 


60 


66 


72 


78 


84 


90 


96 


102 


10S 


114 


120 


126 


132 


138 


144 


150 


7 


14 


21 


2S 


35 


42 


49 


56 


63 


70 


77 


84 


91 


98 


105 


112 


119 


126 


133 


140 


147 


154 


161 


16S 


175 


8 


16 


24 


32 


40 


4S 


56 


64 


72 


SO 


SS 


96 


104 


112 


120 


12S 


136 


144 


152 


160 


168 


176 


184 


192 


200 


9 


18 


27 


36 


45 


54 


63 


72 


81 


90 


99 


108 


117 


126 


135 


144 


153 


162 


171 


180 


189 


198 


207 


216 


225 


10 


20 


30 


40 


50 


60 


70 


80 


90 


100 


110 


120 


130 


140 


150 


160 


170 


180 


190 


200 


210 


220 


230 


240 


250 


11 


09 


33 


44 


55 


66 


77 


88 99 


110 


121 


132 


143 


154 


165 


170 


187 


198 


209 


220 


231 


242 


253 


264 


275 


12 


24 


36 


48, 6 


84 


96108 


L20 


132 


144 


156 
169 


168 


ISO 


192 


204 


216 
234 


228 
247 


240 
260 


252 
273 


264 

286 


276 
299 


288 
312 


300 
325 


13 


26 


39 


52 


65 


7S 


9J 


104 


117 


130 


1431156 


1S2 195 


208 221 


14 


28 


42 


56 


70 


84 


98 


112 


126 


140 


154168 


182 


196210 


224238 


252 


266 


280 


294 


308 


322 


336 


350 


15 


30 


45 


60 


75 


90 


105 


120 


135 


150 


165 


180 


195 


210 225 240|255 


270 


285 


300 


315 


330 


345 


360 


375 


16 


32 


4S 


64 


80 


96 


112 


128 


144 


160 


176 


192 


208 


224 240 256 


272 


28S 


304 


320 


336 


352 


368 


384 


400 


17 


34 


51 


6S 


85 


102 


110 


136 


153 


170 


187 


204 


221 


238255 


272 


289 


306 


323 


340 


357 


374 


391 


408 


425 


18 


36 


54 


72 


90 


10S 


126 


144 


162 


ISO 


198 


216 


234 


252270 


288 


306 


324 


342 


360 


378 


396 


414 


432 


450 


19 


38 


57 


76 


95 


114 


133152 


171 


190 


209 


22 g 


247 


266285 


304 


323 


342 


361 


380 


399 


418 


437 


456 


475 


20 


40 


60 


S0 100'l20 


140160 


180 200 


220 


240 


260 2801300 


320 


340 


360 


380 


400 


420 
441 


440 
462 


460 
483 


480 
504 


500 
525 


21 


42 


63 


84jl05jl26jl47]168|189 


210231 


252 


2 73 


294 


315 


336 


357 


37S 


399 


420 


22 


44 


66 


SS' 1 1 0| 1 32 1 54 1 76 1 98 


220242 


264 


286 


308 


330 


352 


374 


396 


418 


440 


462 


484 


506 


528 


550 


23 


46 


69 


92 115138161 184207 


230253 


276 


299 


322 


345 


358 


391 


414 


437 


460 


483 


506 


529 


552 


575 


24 


4S 


72 


96120144 168192216 


240 264 


288 


312 


336|360 


384 


408 


432 


456 


480 


504 


528 


552 


576 


600 


25 

1 


50 




100425 lso^s^oops 


250 
10 


275 
11 


300 
12 


325 
13 


350375 


400 
16 


425 
17 


450 
18 


475 
19 


500 
20 


525 
21 


550 
22 


575 
23 


600 
24 


625 
25 


2 


3 


"1 


5 


6 


7 


8 


9 


14 


15 



as a is the same as 1 o. When a coefficient occurs just before a paren- 
thesis, it indicates that every term within the parenthesis is to be 
multiplied by that coefficient. 

To multiply with accuracy and rapidity, the product of any two 
quantities, at least from 2 to 12, mint be known at sight. The com- 
binations of these should be practiced until they can ho given correctly 
and without hesitation. 

The following points should also be fixed firmly in mind: Zero 
times any quantity or any quantity times zero is zero. For example, 



16 PRACTICAL MATHEMATICS 

X = 0; X S = 0; 942 X = 0, 6x0=0, etc. One times 
any quantity or any quantity times one gives that quantity as a prod- 
uct. For example, 1 X 7 = 7, 85 X 1 = 85, 1 X 1 = 1, a X 1 =a. 

Table IV gives the product of any two numbers up to 25 X 25. 
To use this table, find one of the two numbers in the upper row and 
the other in the left-hand column. The product of the two will be 
found at the intersection of two imaginary lines drawn parallel to 
the heavy lines shown in the table at 12 X 12 and 20 X 20. For 
example under the 9 and opposite the 8 is found the product, 72. 
That is, 9 X 8 = 72. 

15. Rules for Multiplication, (a) In performing multiplication, 
treat both terms as abstract quantities, alivays using the larger quantity 
as the multiplicand. After the result is obtained, determine from the 
nature of the problem in what units the result should be expressed. 

(b) Place right-hand digit of multiplier directly under right- 
hand digit of multiplicand (with one exception— See Case 2, p. 17). 

(c) Each figure of the multiplicand is multiplied by each signifi- 
cant figure of the multiplier, and the right-hand figure of each product is 
placed under the figure of the multiplier used to obtain it. The sum of 
the several products will be the entire product. When there is a zero in 
the multiplier, multiply by the significant figures only, taking care to 
place the right-hand figure of each separate product under the figure 
used in obtaining it. 

Examples. 1. 

Solution. Having written the multiplier under 
the unit of the multiplicand, multiply the 5 units by 7, 
obtaining 35. Then set down the 5 units directly under 
the 7 and carry the 3; in other words, reserve the 3 tens 

for the tens column. Next multiply the seven tens by 175 multiplicand 
7, obtaining 49, and add the 3 which is carried, and 7 multiplier 

obtain 52 tens (which is the same as 5 hundreds and 2 ]~225 product 
tens). Set down 2 tens and carry the 5 hundreds; mul- 
tiply 1 and 7 and add the 5 which was carried, making 
12, which can be written down in full. 
The product then reads, 1,225. 

2. Find the product of 145 and 13. 

Solution a- It can be seen that 13 = 10 + 3, -iaz y 3 = 435 
hence the product of 13 X 145 will be the same as -, * r y i n = 1450 



(10 X 145) + (3 X 145). Adding the products gives 
1,885 as a result 



1885 



1885 
1246 



PRACTICAL MATHEMATICS 17 

Solution b. The same result is obtained, however, if the 
numbers are arranged as follows: ***° 

Commence with 8, multiply through and write the product 1^> 

435. Under this write the product 1,450 obtained by multiplying ' 435 
by 10. In this latter product the may be discarded but it must ^45 

be remembered to write the 5 under second place. Adding these 
two products, called partial products, gives the final product 1,885. 

3. Multiply 124G by 235. 

Solution. Note that the three partial products are the 

results of multiplying by 5, 3, and 2 where each successive partial ^ J 

product is set one place further to the left than the preceding one. 6230 

Note that is under 0, 8 is under 3, and 2 is under 2; in other 3738 

words, the first figure of each partial product is placed under the 2492 

digit used to obtain it. OQOQin 

Two special cases not covered by the general rules given above 
should be here considered. 

Case 1. When digits of multiplier are separated by ciphers: 
Example. Find the product of 13,456 and 2,004. 

Solution. Although the multiplier contains four digits, in the short 
method only two partial products appear. 

The first figures obtained by multiplying by 4 and 2 appear under these 
respective digits, the zeros simply marking the absence of any other characters 
in the product. 

Regular method Short method 

13456 13456 

2004 2004 

53824 53824 

00000 26912 

00000 26965824 

26912 



26965824 
Case 2. When ciphers are at right of multiplier or multiplicand: 
Example. Multiply 5,760 by 3,000. 

Solution*. In this case it is necessary only to multiply 576 by z'jaQ 
3 giving 1,728; then annex the total number of ciphers found at the Qnfin 

right of both multiplier and multiplicand, in this case four, giving 

as the final result, 17,280,000. 17280000 

16. Short Methods. To multiply by ten and *poivers of ten. 
This method is only a slight variation of that given in the previous 
paragraph. 

* A power of ten is the product obtained by using ten as a factor a certain number of 
Thus, 1,000 and 10,000 are, respectively, the third and fourth powers of 10. 



18 PRACTICAL MATHEMATICS 

Annexing- a cipher to a whole number multiplies that number 
by ten, etc. Thus, to multiply 378 by 1,000 write at once 378,000. 

To multiply by a number a little less than 10, 100, or 1000 the 
process may be shortened as shown in the following: 

Examples. 1. Multiply 254 by 99. 

25400 

Solution. 99 is 100 diminished by 1; hence, multiply 254 ^ r . 

first by 100 and then by 1 and subtract the results. 

25146 

2. Multiply 196 by 997. 

Solution. 997 is 1,000 diminished by 3, hence, multiply 
196 first by 1,000 and then by 3 and subtract the results. 



196000 

588 

195412 



To multiply by 25 annex two ciphers to the multiplicand and 
divide by 4. 

To multiply by 11: 

(a) When the multiplicand contains two figures, place their sum be- 
tween them. If this sum is greater than 10, carry 1 to the third place. 

Example. Multiply 47 by 11. 

Solution. Place 7 in units place. Add 4 and 7. Putting 1 in tens 
place and carrying 1. Place (4 + 1) in hundreds place. Result 517. 

(b) When.the multiplicand is any number, write the right-hand figure 
in units place; then add the first and second, second and third, and 
so on, finally setting down the left-hand figure. Carry as usual. 

Example. Multiply 365 by 11. 

Solution. Write 5; 5 + 6 = 11; write 1 and carry 1; 1 car- 365 

ried + 6 + 3 = 10; write 0; 1 carried + 3 = 4; write 4, making 
the result, 4,015. ~4015 

17. A Method of Checking Multiplication. Add separately the 
figures of the multiplicand and multiplier until they are reduced to 
one figure each ; then multiply these together and again reduce this 
product by addition to one figure. If the multiplication is correct, 
the final result will usually check with the successive additions 
of the figures of the product. 



PRACTICAL MATHEMATICS 19 

Example. Multiply 0,547 by 301 and check. 

Solution. The solution is shown in the 
margin. To check proceed as follows: Taking the 004/ 4 

multiplicand, 6 + 5 + 4 + 7 = 22; 2 + 2 = 4. 301 — 4 

Again for the multiplier, 3 + 1 = 4. ()547 16 — 7 

Their product, 4 X 4 = 16; 1+6 = 7. 19641 

Treating the product in the same manner; q^,,, 7 _ - 
1+9+7 + 6 + 4 + 7 = 34; 3 + 4 = 7. The 1J ' U04 ' ' 
multiplication is correct. 



18. Letters may be used in multiplication as follows: The 
product of a and 6 is a X 6, and the product of a, b, and c is a X b X c. 
If a = 6, b — 4, and c.= 1, the product of a, 6, and c is 6 X 4 X 1 or 
24. The common practice is to omit the multiplication sign when 
letters are multiplied by letters or numbers thus : 2 X xis written 2x; 
2a X 36 is written 6ab; 3a X 46 X 5c is written 60a6c. The order in 
which the letters are placed is immaterial although the alphabetical 
order is usually followed. For example, 60 cab is the same as 60a6c. 

PROBLEMS FOR PRACTICE 
Multiply: 

1. 2,928 by 364. Ans. 1,065,792 

2. 7,319 by 394. 

3. 5,698 by 792. Ans. 4,512,816 

4. 9a by 26. Ans. 18a6 

5. 3a6by4ca\ Ans. 12abcd 

6. 3,186 by 839. Ans. 2,673,054 

7. 42,308 by 692. 

8. 876 by itself. Ans. 767,376 

9. 4a6 by 14c. 

10. 57 by 1,000. 

11. 52 by 99. 

12. 16 by 25. 

13. 92 by 11. 

14. 103 by 25. 

15. There are 746 watts in a horse-power. How many watts 
are there in 20 horse-power? (Suggestion: Multiply 746 by 20.) 

16. A piston has an area of approximately 113 square inches. 
If the steam pressure is 47 pounds per square inch, what is the total 
pressure upon it? Ans. 5,311 pounds. 



> 



20 PRACTICAL MATHEMATICS 

17. The head of a boiler has an area of 11,310 square inches. 
If the pressure per square inch is 40 pounds, what is the total pres- 
sure on the head? Ans. 452,400 pounds. 

18. A concrete mixer delivers 14 cubic yards per hour. A gang 
of men takes away 12 cubic yards per hour. How many cubic yards 
will remain unmoved at the end of 4 hours? At the end of 8 hours? 

19. If there are 12 threads per inch on a screw, how many 
threads are there in 4 inches? 

20. If a piston moves through 468 feet in one minute, how far 
does it travel in 45 minutes? Ans. 21,060 feet. 

21. If a boiler evaporates 1,945 pounds of water in one hour, 
how many pounds will it evaporate in 9 hours? 

22. A certain girder supports 136,925 pounds. How much will 
65 such girders support? Ans. 8,900,125 pounds. 

23. An engine in a certain power plant requires 18 pounds of 
steam per horse-power per hour. If the engine is developing 640 
horse-power, what is the total steam consumption? 

DIVISION 

19. Division is a process of finding how many times one quan- 
tity contains another. In division there are three principal terms, 
the dividend, the divisor, and the quotient or answer. 

The dividend is the quantity to be divided. 

The divisor is the quantity which is divided into the dividend. 

The quotient is the number of times the divisor is contained in 
the dividend. 

When the dividend does not contain the divisor an exact number 
of times, the excess is called the remainder. The remainder being 
a part of the dividend will always be of the same kind as the dividend 
and must necessarily be less than the divisor. 

Division may be indicated in any of the following ways: 24 -5- 2; 

2)24 

Division is the reverse of multiplication, as shown by the follow- 



24 

2 



ing: 



6X7 = 42 42-^6 = 7 42 -7 = 6 

5X8 = 40 40 h- 8 = 5 40 -5 = 8 

20. There are two distinct methods used, viz, long division 



20 
20 



PRACTICAL MATHEMATICS 21 

and short division: in the former all the work is written out but in 
the latter the process is performed mentally and the result only is 
written. Short division is generally used when the divisor does not 
exceed 12. 

The following examples illustrate the two processes. 

Examples. 1. Divide 720 by 5. 

Lone: Division 
144 quotient 

Solution. In long division it is found that 5 is 5)720 

contained in 7 once. "Write 1 as the first figure of the J^_ 

quotient, and subtract giving a remainder of 2. To 22 

the remainder *annex the next figure of the dividend, 20 

and divide as before, obtaining 4 as the second figure 
of the quotient. Annex which is the next figure of the 
dividend, and divide again by 5, obtaining 4 as the last 
figure of the quotient with no remainder. The division 
is now complete. Short Division 

5 )720 

144 quotient 
It often happens, after bringing down a figure from the dividend, 
that the number is too small to contain the divisor. In this case 
place a zero in the quotient, and continue bringing down the figures 
from the dividend until the number thus formed will contain the divisor. 
The following example illustrates this point: 

2. Divide 10,426 by 13. 

Solution. It is seen that the first two places gQ2 

in the dividend are less than the divisor; therefore, iQ V1fU9f ' 

three places must be taken. 13 is contained in 104 int 

exactly 8 times, giving 8 as the first figure in the quo- *-®* 

tient. There being no remainder, the next figure of 20 

the dividend, when brought down, stands alone and 26 

is, of course, less than 13. Place a zero as the second 
figure of the quotient and annex the next figure 6 of 

the dividend making 26, into which 13 goes exactly twice. This makes the 
complete quotient 802 without a remainder. 

21. Rules for Long Division, (a) Write the divisor and dim- 
dend in the order named, and draw a cursed line between them. 

(b) Fi nf l how man j/ times the divisor is contained in the left-hand 

* In the above solution "mnox", moaning to place after, is used. The opposite 

term "prefix", meaning, to pi !. Be careful to make a distinction 

between "annex" and "add". 



22 PRACTICAL MATHEMATICS 

figure or figures of the dividend, and write the number in the quotient 
over the dividend. 

(c) Multiply the divisor by this figure of the quotient, writing the 
product under that part of the dividend from which it teas obtained; 
subtract, and to the remainder annex the next figure of the dividend. 

(d) Find how many times the divisor is contained in the number 
thus formed, and write the figure denoting it at the right hand of the last 
figure of the quotient. 

(e) Proceed in this manner until all the figures of the dividend 
are divided. If there is a remainder after dividing all the figures of the 
dividend, place the remainder over the divisor with a line between them, 
and annex to the quotient. 

(f) The proper remainder is, in all cases, less than the divisor. 
If, in the course of the operation, it is found to be larger than the divisor, 
th is indicates that there is an error in the work and that the figure in the 
quotient should be increased. 

Example. Divide 5,441 by 26. 
Solution: 2 Q9 t V 

26)5441 
52 
241 
234 



7 remainder 

Proof. In order to prove division, multiply the quotient by the 
divisor, and add the remainder, if there is any. If the quantity thus 
obtained gives the dividend, the work is correct. 

22. *Short Methods. To divide by ten and powers of ten. 
From the right in the dividend point off as many places as the divisor 
contains ciphers. The figures so cut off represent the remainder, 
to be written over the divisor and annexed to the quotient. 

To divide by multiples of ten. 

Example. Find the quotient of 18,653 -^ 3,000. 

Solution. Mark off as many figures in the divi- 553 

3000 



dend as there are ciphers in the divisor, thus dividing 



by 1,000. This leaves 18 to the left of the mark to be 

divided by 3, giving a quotient of 6 with a remainder o|UUU)lo|00o 

of 653. Place this remainder over the divisor and 18 

653 pro 

annex it to the 6, giving, finally, 6 ittt/jtj- ^ 

♦These methods may be made more comprehensive and valuable after the subject of 
Decimals has been discussed. 






PRACTICAL MATHEMATICS 23 

To divide by 25. 
Multiply the dividend by 4 and divide the product by 100, by cutting 
off two figures to the right. 

2o. Letters may be used in division as follows: a divided by 

'■■ may be expressed as a -s- b or -■-.- a divided by c, as a 4- c, or -• 

If a = 6 and 6 = 3, a divided by b will equal 6 ■*■ 3 = 2. If a = 
and c = 2, a divided by c will equal 6 -s- 2 or 3. In dividing terms 
containing letters, it is useful to remember the statement on page 
20, that division is the reverse of multiplication. For example, Aabc 

divided by 2c equals 2(d), for 2c X 2ab equals -iabc. 

Examples. 1. Divide IQabcd by 4ad. 

Solution. The coefficient of the divisor is ^. oc 

contained in the coefficient of the dividend 4 times. 
i 4 in the quotient. Again a and d of the divisor 
are contained in (7 and d of the dividend exactly once 
but ad is contained in abed, be times for ad X be equals 
a } >ed. The complete quotient is, therefore, 46c without 
a remainder. 



4ad)16abcd 

IGabcd 



2. Divide (I5abc + (jab) by 3b. 

tiox. In this problem the two terms of r i o a 

the dividend are divided separately and the two quo- — - — — r 

tients connected by the proper sign. 15abc divided ou)Loaoc -f- bab 

by 36 gives a quotient of oac; 6a6 divided by 36 gives locibc -f- bob 
2a as a quotient. The final quotient is, therefore, 
oac 4- 2n. 

PROBLEA\S FOR PRACTICE 
Divide: 

1. 65,814 by 6. Ans. 10,969 

_ 6a5 by 36. Ans. 2a 

3. 3,870 by 10. 

4. 24abcd by <V. Ans. 4abc 

5. 9,473 by 100. Ans. 94 T V* 

6. 13,987 by 1000. 

7. dOabcd by one. 

8. (18abc+12abd) by 36. 

9. If eoal costs 5 dollars per ton, how many tons may be 
bought for 275 dollar-? (Suggestion: Divide 275 by 5.) 



24 PRACTICAL MATHEMATICS 

10. A steamer runs 270 miles in 23 hours. What is her average 
speed per hour? (Suggestion: Divide 276 by 23.) 

11. How many 30-foot rails are there in a double track (four 
rails) two miles long? There are 5,280 feet in one mile. (Sugges- 
tion: Divide 2 X 4 X 5,280 by 30.) 

12. If 9 days' work will pay for 6 tons of coal at 6 dollars per 
ton, what is the price of a day's labor? 

13. A company furnishes equal power to 26 establishments. 
The total horse-power is 8,450. How much does each receive? 

14. If a locomotive goes to shop for inspection after every 
average run of 283 miles, how many times would it be in shop during 
runs aggregating 10,188 miles? Ans. 36 times. 

15. A certain boiler supplies steam for heating. If there are 
180 square feet of heating surface in the boiler and each radiator 
requires 12 square feet of heating surface of the boiler, how many 
radiators can be supplied by the boiler? Ans. 15 radiators. 

16. If 800 cubic feet of air are required for each person, how 
many people can occupy a room that contains 21,600 cubic feet? 

17. If 1 foot of 1 inch pipe is allowed for every 90 cubic feet of 
space in heating a factory, how many feet of the same pipe will be 
required to heat 225,000 cubic feet of space? 

FACTORING 

24. An integer is a whole number, as 1, 8, 15. 

All numbers are either odd or even. An odd number is a num- 
ber that cannot be divided by 2 without a remainder, as 3, 9, 13, etc. 
An even number is one that can be divided by 2 without a remainder, 
as 4, 6, 8, etc. 

A prime number is a number which can be exactly divided only 
by itself or 1, as 1, 3, 5, 7, etc. 

A factor of an integer is a number which is contained an exact 
number of times in the given integer. 

A prime factor is a factor which is a prime number. 

Numbers are prime to each other when they have no factor in 
common. Thus 7 and 11 are prime to each other; also 18 and 25, etc. 

In speaking of the factors of a number it is not customary to 
include the number itself or 1. For this reason a prime number is 
said to have no factors. 






PRACTICAL MATHEMATICS 25 

Example. What are the prime factors of 10? 

Solution. Divide 1G by 2 which is the least prime number 

greater than 1, and obtain S as the quotient. Since S may again be 2)16 

divided by 2, the division is carried out giving 4 as t he next quotient. oY"^ 

Next divide 4 by 2 giving a quotient of 2, which is a prime number I — 

and cannot be further divided. Xow the several divisors and the ~z__ 

last quotient, all of which are prime, are the prime factors of 16. 2 

The rule for finding prime factors may be expressed as follows: 

Divide the given number by the least prime number (greater than 
1) that will divide it; and divide each quotient, in the same manner. 
Continue dividing until a prime number is obtained as the quotient. 
The several divisors and the last quotient will be the prime factors 
desired. 

Example. What are the prime factors of 78? 

Solution. 2)78 

S)39 
13 Ans. 2, 3, and 13 

25. Factoring by Observation. Certain numbers have such 
characteristics that some of their factors may be found without the 
trouble of an actual operation in division. The simplest of these 
rules for factoring by observation are as follows: 

2 is a factor of all even numbers. 

3 is a factor of all numbers the sum of whose digits is divisible by 3. 

4 is a factor of all numbers whose last two places are divisible by 4- 

5 is a factor of all numbers ending in 5 or 0. 

10 is a factor of all numbers whose last figure is 0. 

Examples. 1. Factor the number 56. 

Solution. By inspection it is seen that 56 4)56 

is divisible by 4, and although 4 is not a prime oTTZ 

factor, its use saves one division. The result- 1 — 

ing factors are 4, 2, and 7 7 Ans. 4, 2, and 7 

2. Factor the number 720. 

Solution. According to the rule in §25 
this number is divisible by 10 and, although 10 ) /^i) 

10 is not a prime number, the process of factor- 4)72 

ing is shortened by its use. The quotient 72 2)18 

ia divisible by 4, which should be used instead — 

of dividing twice by the factor 2. The remaining °' 

factors are 2, 3, and 3. 720 may also be divided 3 Ans. 10, 4, 2, 3, 

into the factors 10, 8, and 9, the two latter being an( J 3 

equal, respectively, to 4 X 2 and 3X3. 



26 PRACTICAL MATHEMATICS 



3. Factor the number 42G. 




Solution. 


2)426 




3)213 



71 Ans. 2, 3, and 71 
Find the prime factors of the following: 

1. 36 Ans. 2, 2, 3, and 3 

2. 30 

3. 68 

4. 148 

5. 189 Ans. 3, 3, 3, and 7 

6. 1,264 

7. 2,552 

8. 1,932 

9. 91 

10. 2,508 Ans. 2, 2, 3, 11, 19 

CANCELLATION 

26. Cancellation Methods. When a series of multiplied factors 
is to be divided by a second series, the operation may be shortened 
by a process called "Cancellation". The first series is placed above 
and the second below a horizontal line and divisions are performed 
between the factors on opposite sides of the line. Any number on 
one side which is exactly divisible by a number on the other side of 
the line may be so divided. 

Examples. 1. Divide 18 X 5 by 10 X 3. 
Solution: .Arranging the first series above and the second 
series below the line we have 

6 6 1 

18 X 5 = 1$ X 5 = 1$ X $ 

10X3 10X3 Z0X3 

1 2 1 

It will be seen that 18, the first term above, and 3, the second term 
below the line, are divisible by 3. The 18 is, therefore, crossed out 
or ' 'cancelled out" and a 6 placed above it; the 3 below is also can- 
celled and a 1 placed below it. Similarly 5 is contained in 5 and 
10, the remaining factors, and these should also be cancelled out as 
indicated, a 1 being placed above the 5 and a 2 below the 10. 

Further inspection shows that the 6 which remains above the 
line is divisible by the 2 which remains below the line and this can- 



PRACTICAL MATHEMATICS 27 

ccllation gives a final result of 3. The complete process is 

3 

1 
ZBX5 = 3 ^ 3 

X0 X 3 1 
2 1 
1 

6 X 8 X 12 X 18 X 24 = ? 
2X3X2X4X6 

Solution*. To cancel, select 
any two numbers one above and one 
below the line, and find some num- 
ber that will divide each of them 
without a remainder. For instance, 

it is seen that 2 will be contained an 3 4 4 3 6 

exact number of times in both 6 and 2. 0X$XZ2XX$X2^ 

Then perform the division, crossing 9 V S? V 9~V 4~V fl = 

out both numbers and placing the re- 1 1 1 -■ 1 

suits directly over and below, the 

numbers crossed out. Proceed in this 3X4X4X3X6 
manner until there is no longer a ; = 864 

number below the line that can be 
cancelled with one above the line. 
Then multiply together all the num- 
bers above the line and use as a divi- 
dend; and multiply together those 
below the line, and use as a divisor. 

3 16 X 2 X 15 X 4 = 
32 X 6 X 8 X 22 

Solution. In an example where the 
product above the line, after cancelling, is 
less than the product below the line, the 
result is allowed to stand as obtained, thus: 
Since 5 and 22 do not cancel into any of the 
other numbers the result is the product of 
the quantities in the numerator divided by 
the product of the quantities in the de- 
nominator. 

4. Four men have the task of sorting 24 boxes of castings, each 
box containing 720 pieces. If each man can handle 12 pieces per 
minute, how many hours will it take the men to sort the castings? 

Solution". A little examination of this problem will show that the total 
number of pieces to be handled will form ilu- quantity above the line, that is, 24 



115 1 

u x % x n x i 


5 


32 X X $ X 22 
2 3 2 

1 


88 





1 






1% 




6 


n 




U X 720 




i 


X 12X00 


1 


1 




1 



28 PRACTICAL MATHEMATICS 

times 720. As there are 4 men and each can handle 
1 2 pieces per minute, the number handled per minute 
will be 4 X 12 and the number per hour will be 
4 X 12 X 60. This quantity is placed below the line 

as the total number of pieces divided by the number ' ^ . , nr± ~ 6 

handled per hour will give the required result. Carry- 
ing on the cancellation as before, the final result is 6, 
that is, it will take the men 6 hours to sort the castings. 

Rules. Cancel the common factors from both dividend and divisor. 

Next, divide the product of the remaining factors of the dividend 
by the product of the remaining factors of the divisor. 

It is seen, therefore, that cancellation is merely a combination 
of the processes of factoring and division. 

PROBLEMS FOR PRACTICE 

Divide : 

1. 2X3X8X12X24 by 6X4X36X4. Ans. 4 

2. 18X24X32X36 by 9X48X4X18. 

3. 15X20X25X27 by 15X18X25X10. Ans. 3 

4. 40X48X54X60 by 30x24x72x3. 

5. 12X60X36X70 by 28X5X48X6. Ans. 45 

6. 32X36X33X45 by 24X30X44X9. 

7. 18 piers, each consisting of 5 piles, were set by 10 men in 3 
days. What is the cost of driving each pile if each of the men receive 
3 dollars a day? 

8. Each of 40 teamsters hauls 9 yards of sand per day for 4 
days at 1 dollar per yard. How many men with wheelbarrows 
will earn the same amount in two days wheeling 3 yards per day at 
1 dollar per yard? Ans. 240 men. 

REVIEW PROBLEMS 

1. A steam plant has two engines of 934 horse-power each. 
The starboard engine of a steamship develops 3,218 horse-power 
and the port engine 3,232 horse-power. How much greater power 
is developed in the ship than in the steam plant? Ans. 4,582 H. P. 

2. A pump delivers 33,720 cubic feet of water in two hours (60 
minutes each). How many cubic feet are delivered per minute? 

Ans. 281 cu. ft. 

3. A nine-foot blower makes 175 revolutions per minute. How 
many revolutions will it make in 29 minutes? Ans. 5,075 rev, 

4. There are 360 rivets in a hundred pounds. How many 



PRACTICAL MATHEMATICS 29 

pounds will be required to afford 04,800 rivets. Ans. 1S,000 pounds. 

5. The heating surface in a locomotive is 88 square feet in the 
fire box and 702 square feet in the tubes. What is the total heating 
surface? How many times as much surface is there in the tubes than 
in the fire box? . ^ 880 square feet 

|0 times 

6. The weight of a battery of eight marine boilers and their 
equipments is as follows : 

Boilers complete with mountings .295 tons 

'Water in boilers 73 " 

Funnels 50 " 

Stoke-hole plates, floors, etc 23 " 

Feed pumps 7 " 

Fans and fan engines 8 " 

Feed regulators 2 " 

Tools and fittings 2 " 

Spare gear 10 " 

'What is the weight of two such batteries? 

7. At 100 degrees Fahrenheit a cubic foot of water weighs 62 
pounds. At 205 degrees a cubic foot weighs 60 pounds. What 
is the difference in weight between 173 cubic feet of water at 100 
degrees and 189 cubic feet at 205 degrees? Ans. 614 pounds. 

8. A roof is composed of 11 frames. The weight of one frame 
in detail is as follows: 

2 rafters each weighing , 875 lbs. 

5 rods, each weighing 176 " 

16 bolts, each weighing 5 " 

8 bridle-straps, each weighing 15 " 

2 piers supporting rafters at ridge, average each. . . 11 " 

6 pieces at foot of struts, average each 11 lbs. 

4 pieces uniting rafters at junction in strut, with 

bolts and nuts, each 44 

2 rafter shoes, each 144 

2 cast-iron struts, each 1 54 

What is the total weight of the roof? Ans. 40,590 pounds. 

9. A double mine ventilating fan runs at the rate of 84 revolu- 
tions per minute. It gives 2JS18 cubic feet of air per revolution. 



30 PRACTICAL MATHEMATICS 

How many cubic feet should it give per minute? 

Ans. 236,712 cubic feet. 

10. If oak-tanned leather belting costs 2 dollars per foot, and 
four-ply stitched canvas belting costs 1 dollar per foot, what is the 
difference in the cost of 40 feet? Ans. 40 dollars. 

11. If there are 9,326 heat units in a pound of lignite coal, how 
many heat units in 287 pounds? 

12. If 23 square feet of No. 30 sheet iron weighs 11 pounds, 
how much will 23 square feet weigh if three times as thick ? 

13. The weight of the masonry of a bridge and an engine passing 
over it is 1,698,575 pounds. The engine weighs 198,560 pounds. 
What is the weight of the masonry? 

14. A pound of Pennsylvania petroleum will theoretically 
evaporate about 22 pounds of water. How many pounds are neces- 
sary to evaporate 4,378 pounds of water? 

15. A cubic foot of hemlock weighs 25 pounds. A cubic foot 
of iron weighs 450 pounds. Find the difference between the weight 
of 230 cubic feet of hemlock and 87 cubic feet of iron. 

16. Three guy ropes are fastened to a stake. The pull on one 
rope is 560,800 pounds; the pull on the second is 118,421 pounds; 
and on the third is 104,863 pounds. What is the total pull? 

17.* Divide (20ac X 4bd) by 166. Ans. 5acd 

18.* Divide the product of la and 6bcd by Sad. 

19. Let a, b, and c have, respectively, the values 4, 7, and 2; 
Subtract a from the sum of b and c and give the numerical result. 

Ans. 5 

20. Give values of 5, 9, 6, and 12 to a, b, c, and d, respectively; 
find the numerical result of adding 3a, 7c, and 5d, and subtracting 
86 from this sum. 

21. The record of a pieceworker in a shop for one week was as 
follows: for Monday 122 pieces, Tuesday 140, Wednesday 114, 
Thursday 154, Friday 132, and Saturday 142. What is his aver- 
age day's work. (Suggestion: An average is found by dividing 
the sum of the various items by the number of items.) 

Ans. 134 pieces per day. 

* Work this problem by cancellation. 



PRACTICAL MATHEMATICS 

PART II 



FRACTIONS 



27. A fraction is an expression denoting one or more equal 
parts of a unit, and may be regarded as indicating division. The 
fraction is written, i. e.* the division is indicated, by placing the 
dividend over the divisor with a line between. 

Thus, — denotes that 5 of the 12 equal parts of a unit are to be 

taken. — - is merely a different way of writing 5 -5- 12, and the 

fraction is used because the dividend is less than the divisor. 

To S ay — of a 100 pound keg of bolts indicates, that the keg of 

bolts is to be divided into 4 equal parts, and 1 of these parts taken. 
In this use, the unit may be considered as made up of a number 
of equal parts, but when used separately and without reference to 
any certain thing, it is convenient to consider the fraction merely 

1 Q O""" 

a> an unperformed division; thus,— =1-?- 4; — =3t4; — = 

&75 -^ 361. 

The quantity below the line is called the denominator. It shows 
into how many equal parts the unit is divided. 

The quantity above the line is called the numerator. It shows 

how many of these equal parts are taken. Thus, in the fraction --, 

the 12 -hows that the unit has been divided into 12 equal parts, and 

the 1 shows that one of the 12 parts is taken. In the fraction— the 

unit i- divided into t> parts and a of these parts arc taken. 

The numerator and denominator are called the terms of a 

fraction. 



The letters "i. e." are u 



32 PRACTICAL MATHEMATICS 

A fractional unit is one of the equal parts into which the unit in 
question is divided. Thus, if the unit is divided into fourths, 

— is the fractional unit; if divided into b parts, —is the fractional 

unit. 

Since any number divided by 1 gives a quotient equal to the 

dividend, any whole number may be expressed as a fraction by writing 

4 4 

1 for the denominator. Thus, 4 may be written — , as — = 4 -*■ 1 = 4. 

The value of a fraction depends upon the value of the fractional 
units and the number of these units taken, or simply upon the division 

4 
of the numerator by the denominator. Thus, in — , the quotient 

of 4 -r- 2 is 2, and the value of the fraction can be expressed 
as 2. 

If the numerator and the denominator are equal, the value of the 
g 
fraction is 1. Thus, -—- may be expressed as 8 -t- 8 and is equal to 

8 

1. This shows that one unit is divided into eight parts, each part 
being an eighth and that 8 of these are taken, making a unit 
or 1. 

Strictly speaking a fraction is less than a unit; hence if the numera- 
tor is less than the denominator, the value is less than 1, and it is 

known as a proper fraction. For example, the expression — means 

that a unit has been divided into nine parts, each part forming a 

fractional Unit having a value of — , and that eight of these parts are 

taken. This is one less than the nine parts necessary to make a 

o 
unit, and therefore — is less than 1. 

y 

If the numerator is greater than the denominator, the value of 

the fraction is greater than 1, and it is called an improper fraction. 

o 

Thus, the fraction — is an improper fraction, because 7 is contained 

once in 8 with a remainder, or, expressing it in another way, because 
eight parts, each one being a seventh of a unit, have been taken, form- 
ing a unit and one seventh. 



PRACTICAL MATHEMATICS 33 

REDUCTION OF FRACTIONS 

28, To reduce a fraction is to change its form without changing 
its value. 

To reduce a fraction to higher terms multiply both numerator and 

denominator by the same quantity. Thus, -j- = =7^- The 

4 4 X o 1 2 

value oi the fraction has been increased three times by multiplying 
the numerator by 3, and then decreased just as many times by mul- 
tiplying the denominator by 3. If in the fraction — - both the 
1 • J d 

numerator and denominator are multiplied by c, the value of the 

be 
fraction is not changed but the form is changed to — • - Thus, 

multiplying both numerator and denominator by the same quantity 
does not change the value of the fraction. 

To reduce a fraction to lower terms divide both numerator and 

4 — 22 
denominator by the same number. Thus, - — '—— - = — . In this 

l) "5" A o 

case dividing the denominator by 2 changes the fractional parts from 
sixths to thirds, which are twice as large as sixths and this much of 
the operation has doubled the value of the fraction. Dividing the 
numerator by 2 decreases the number of parts to one-half of the 
original number. Therefore dividing both numerator and denom- 
inator by the same quantity does not change the value of the fraction. 

In the >ame way — may be reduced to ^— by dividing both numer- 
' mn ' n 

ator and denominator by m. 

A fraction is reduced to its lowest terms when its numerator 
and denominator have no common factor other than 1; that is when 

. , rp, 1 2 15a 

the terms are prime to each other. 1 hus, ~— ,—-,--,-— , are 
1 2 3 10 6 

4 
reduced to their lowest terms, but — is not, as 1 and 6 may both 

.2 ad 

be divided by 2, reducing the fraction to - 1 and - is not, as ad 
J 3 ac 

and "<• may both be divided by a, reducing the fraction to — 

c 



34 PRACTICAL MATHEMATICS 

An entire quantity is one which has no fractional part as 3, 11, 
orb. An entire quantity, which is also a number, is called an integer. 
For example, 3 and 11 are integers. 

A mixed quantity consists of an entire quantity and a fraction. 

1 7 ' h 

Thus, 5—, 3 — , anda + — are mixed quantities. To reduce a 
6 8 d 

mixed quantity to an improper fraction, multiply the entire quantity 

by the denominator of the fraction, add to this product the numerator 

of the fraction and place their sum over the denominator. To reduce 

3 — to an improper fraction, reduce 3 units to eighths by multiplying 

8 

3 by 8 obtaining 24; add this to 7 and the result, 31, is the numerator 

of the fraction. 3— = — . In the same way a + — = — 

8 8 d d 

To reduce an improper fraction to a mixed quantity divide the 

numerator by the denominator, write the quotient as the entire 

quantity of the mixed quantity and place the remainder over the 

divisor (denominator of improper fraction) as the fraction. 

31 
Examples. 1. Change — to a- mixed number. 

z 1 

Solution. Divide the numerator 31 by the ]* ° quotient 

denominator 21, and place the remainder 10 over the 
divisor 21 as the fraction of the mixed quantity. As 
21 twenty-firsts constitute one unit, 31 twenty-firsts 
constitute one unit and ten more twenty-firsts. 10 remainder 

2. Reduce to a mixed number. 

16a 

Solution. Following the process already given 

35a6 _, .36.- , tl ,, Sab , 36 

-— - — = 26 + — r, tor by cancellation — — becomes-^, 
16a 16 16a 16 

36* 
giving the final form 26 + -h-. This quantity 26 -f- 

3& 26+ f i quotient 

— can be changed to the original quantity thus : Fol- iCnYViah 

lowing the rule given above, 16 X 26 equals 326 and 32a6 

adding to this 36 and placing 16 as the denominator ~ 3a6 remainder 

356 
we have -y— . Multiplying both numerator and denom- 
inator by the same quantity a does not change its 

value and gives — - — , the original fraction. 
16a 



21)31 
21 



*This must not be expressed as 26 — , as this would mean 26 X — <■ 

lo 16 



PRACTICAL MATHEMATICS 35 


PROBLEMS FOR PRACTICE 


Reduce to mixed quantities: 


Change 


to improper fractions: 


25679 




4. 


3| 




2. l 4 




5. 


90— 
9 


3. *2 An, 234- 
5 o 




6. 


21^- 

20 


Reduce to lowest terms: 








- 369 
936 




8. 


288 
360 


Find the proper numerators: 








1 ? 3 
9. — = — 10. — = 

16 96 5 


? 
= 525 




n 21 ? . 441 
1L 22 = 462 AnS -462 



Suggestion: Problem 9. 96-^16 = 6. Hcnco, 6 is the multiplying factor 
for numerator and denominator of fraction. See §28, second paragraph. 

Reduce to mixed quantities: 



12 Uah0 \ns ^/+ 4a 


13. 


16a 


DOC 5 


6 


Reduce to lowest terms: 






-. i 8a . 2a 
14. - Ans. y 


15. 


166c 
32c 


Change to improper fractions: 






16. 4 4- 6 Ans. 2S ^ +G 
lg S \> 


17. 


,- ' 3a 
,« + - 


Find the proper numerators: 






18- i = ^ 

5a loa 


19. 


5 ? 
66 126 



* 10- 
AnS ' T26 

LEAST COMMON DENOMINATOR 

29. The product of a number of factors is called a multiple 
of any one of them. Thus, 6 is a multiple of 3; 21 is a multiple of 
7; and ab is a multiple of 6. 

Now the denominators of any group of fractions must be reduced 
to some common multiple, called a common denominator, before the 
fractions may be added, subtracted, or compared in value. This 



36 PRACTICAL MATHEMATICS 

common denominator may be obtained by multiplying together the 
denominators of the fractions. Suppose, for example, it is desired 

to reduce -~- and — of a foot to fractional parts of a foot, having a 

common denominator. Let the common denominator be 3 X 4 = 12. 

To reduce— -to higher terms, i. e., to a fraction having the given 

o 

denominator 12, multiply both numerator and denominator by the 

14 . 1 . 3 

factor 4; thus, — = — . In a similar manner —-is found to be — -. 

The least common denominator of several fractions is the least 
quantity which may be divided by each denominator without a re- 
mainder. Thus, 24 is the least common denominator of the fractions 

5 3 7 

— , — , and — -, and ab is the least common denominator of the 

6 8 12 

fractions — and — . 
a b 

The least common denominator of several fractions contains all 

of the prime factors of the given denominators. Thus, either 96 

5 7 

or 192 is a common denominator of the fractions — - and — , and 

x£» 10 

contains all their prime factors, 2, 2, 3, and 2, 2, 2, 2. The least 
common denominator, however, is neither 96 nor 192, but 48, because 
48 is the least quantity which contains the prime factors of both 12 
and 16 the greatest number of times that they appear in either one of 
these quantities. 

12 = 2 X 2 X 3 

16 = 2X2X2X2 

48 = 2X2X2X2X3 
Since 2 is a common factor of 12 and 16 and appears four times in the 
latter, it is taken four times. There are no other common factors 
and therefore, the four 2's and the remaining factor 3, are all the 
factors which make up the least common denominator, and their pro- 
duct is 48. 

Therefore, to find the least common denominator, separate the 
denominators into their prime factors and take each factor as many 
times as it appears in the denominator containing it the greatest 
number of times. 



PRACTICAL MATHEMATICS 37 

Examples. 1. Kind the leasl common denominator of the 

fractions, - . - ,— . and — ,. 

Solution. First, write the denominators in a 
row. as shown in the margin. Now the least quantity 
to contain 3, J. 9, and 12 must be the smallest quantity 

that will contain the factors of each of them, hut no 
other factors. Then all the prime factors that the com- 
mon denominator contains must he found. 2 is a prime 
factor of 4 and L2. Therefore, it must he a factor oi 
any quantity that contains 4 and 12 without a re- oy>. , 0# 10 

mainder. Divide the 4 and 12 by 2, writing the quotients - '-'— ! '— -- 

hclow. carrying down the 3 and 9 which are not divisible - ' ' ~*» ■'*» b 

by 2. Again it is seen that 2 is a factor of 2 and 6, ;', .1; 1; 9; 3 

and the operation i- repeated, obtaining 3, 1. 9, and 3. i . i . •). 7" 

dividing by 3, the result is 1, 1, 3, and 1. Now ' -' 

final quotients have no common factor, and must 
be factors of the least common denominator just as 2, 
2. and :> are. Disregarding the l's, 2X 2X3 X 3 = • 
The result 36 is the leasl common denominator. When- 
ever I'b appear, they may be disregarded, as mul- 
tiplying by the factor 1 produces no change in the 
quantity. 

2. Find the least common denominator of the fractions 

5 



1 2ax 



and 



2)12a; 


48%; 426 


:i) 6a; 


24by; 2\b 


2) 2a; 


Sby; lb 


b) la; 


4by; lb 



426.x 



4sby 42bx 

Solution-. Dividing through by the sue- x)12ax;4Sby; 

cessive factors, remainders are a, 4y, and 7. There- 
fore, *X2X3X2XbXaX4/yX 7 = 
SSQabxy. The work can sometimes be shortened 
by dividing out any factor common to all the de- 
nominators even though it is not prime. It may 
be observed thai 6 is such a common factor, and 
could li.r ised as a divisor in place of 2 a * V > 

and 3. 

Sometimes the least common denominator may l>r found by 

15 } 

inspecting the fractions. In . • -, and _ , since 12 is a multiple 

. and any quantity which has 12 as a factor must contain 6, it 
is only accessary to find the least common denominator of the deno- 
minators 12 and 7, which i> 84. 

7 r< fractions to their least common denominator, first reduce 

each fraction to its lowest terms, and then find the least common 



38 



PRACTICAL MATHEMATICS 



denominator of these fractions. Each fraction must now be changed to 

a fraction whose denominator is the least common denominator, which 

operation, it is readily seen, is merely the reducing of the fractions 

to higher terms. Eor example, in the preceding paragraph, the least 

15 4 . 

common denominator of the fractions — -, -— , and — - , is found to 

6 12 7 

1 14 5 

be 84; when the reduction is accomplished, — becomes — — ; — ■ be- 

oi 1 Z 

3 5 4 48 

comes -— ; and -— becomes — -• 

84 7 84 



Examples. 1 . Let — -, — , ~ 

18 4 9 



9. 7 

, and — , be reduced to equivalent 



12 



fractions with a least common denominator. 

Solution, rg must be reduced to lower terms by dividing both numera- 
tor and denominator by 6, which changes the form to |. The least common 
denominator of £, |, f, and T 7 g- has just been found to be 36. 

1 12 



36 ^ 3 = 12 
36 4- 4 = 9 
36-i- 9 = 4 

36 -5- 12 = 3 



12 X 1 = 12 

9 X 3 = 27 
4X2 = 8 
3 X 7 = 21 



3 


36 


3 
4 


27 
"36 


2 
9 : 


8 
"36 


7 
L2 = 


21 
~36 



15 7 5 

2. Reduce -wp. — > 7777- >■ and -r^r- to equivalent fractions with a 
36 arc 486?/ 426rc n 

least common denominator. 



15 

Solution. The fraction — — may be reduced by dividing both numera- 

15 5 

tor and denominator by 3. — — = — - — • The least common denominator 

5 7 5 

of vt^— . ttt; — i and tnl was found on Page 37 to be SSGabxy. 
12ax' 486?/ 426x 



SSQabxy -f- \2ax = 28a?/ 
336abxy 4- 48a?/ = 7 ax 
SZGabxy -s- 426a; = Say 



286?/ X 5 = 1406?/ 
lax X 7 = 49a^ 
Say X 5 = 40a?/ 



5 1406,?/ 

12ax 336abxy 

7 49arc 



486?/ SSQabxy 

5 40a?/ 

426a: = S36abxy 



PRACTICAL MATHEMATICS 39 



PROBLEMS FOR PRACTICE 

1. Change— and— to fractions having a common denomi- 
nator. 

4 7 5 

2. Change -— -j -— -, and — to fractions having a common de- 

5 12 6 

nominator. 

3 5 2 1 

3. Change — , — ,-^-, and — to fractions having a common 

/ o o 2 

denominator. 

Change to common or improper fractions using the least common 
denominator: 

4. 7T- tV ih ancI tV 1°- fi tV and £. 

5. 5J, 2|, and If. 11. A an d — 

o c 

6. f tV 3}, and 7. 12. ±, A, an d JL 

7. 4, }> t» and f- 13 - — » — » — ,and— . 

m n mp np 

8. J, f \, and T 4 T . 14. — , — ,and — . 

xy xz yz 

9. A, fi i> an d 4- 15. — , — ,and 

yz xy y 

ADDITION OF FRACTIONS 

30. To add together two or more fractions, the fractional units 
must be of the same size; in other words, they must be reduced to a 
common denominator before the addition can be accomplished. 

For example, suppose it is desired to add the fractions — and — 

These may for convenience represent — and — of a foot. Now, it 

4 o 

is well known that a foot contains 12 inches, and the — mav be con- 

4 

3 1.4 

sidered a^ 3 inches or — of a foot, the — as 4 inches, or -— of a foot. 



10 



PRACTICAL MATHEMATICS 



The required sum is then 3 + 4=7 inches, or j = — 

1 ' 12 T 12 12 

This process is graphically illustrated in Fig. 1. 



feet. 




Fig. 1. Graphical Representation of Addition of Fractions 

Now if an example be taken in which there are more than two 

5 7 

fractions, it will be noted that the procedure is the same. — -j (- 

8 12 

11 3 

r^r 4- -j- = ? It may be seen by inspection or by the process already 

given that 24 is the least common denominator of the several fractions. 
The next step is to change the numerators of the fractions so that 
they will express the same value with the common denominator 24 

5 15 

as they now express with their respective denominators. — = — - 

tt: = ^rr- 7T-r will stand as written: -— - = -—-. Adding the several 

12 24' 24 4 24 8 

numerators and placing the sum over the common denominator the 

r o 

required sum is found to be—, which is an improper fraction. Re- 
ducing the improper fraction to a mixed quantity gives 2^-f, or finally 

9 5 

Again, to add - — , — - 3 and — — ' first find the least common de- 
2p 5p Sp 

. . , . , . _ m, d 15d d 6d d lOd 
nominator, which is SOp. lhen — = — — ,* •— - = t^t—',-^- = -rr— . 
r 2p 30p bp 30p 3p 30p' 

d d_ d_ 15d §d_ lOd ZU 

2p + 5p + 3p ~~ 30p + [30p + 30p " ~30p ' 

If the problem involves letters only, the process is essentially 

the same. If — and — are to be added, they must first be reduced to 

n p J 



PRACTICAL MATHEMATICS 



ap _ b 



bn 
np 



41 

a h 

. , + _ 

n p 



their common denominator, np, — 

1 /i 

= \- As the relation between ap and bn is not known, the 

7?p Tip z 

sum of the numerators can only be represented by the expression 

ap -f bn 



ap + 6/?, giving the final sum 



np 



To add mixed quantities, add the entire quantities and the frac- 
tions separately, and if the fractional result of this addition is an 
improper fraction, reduce it to a mixed quantity and add the entire 
part to the sum of the entire quantities already obtained. 

Examples. 1. 20f + 13J + 7| - ? 



Solution. Adding the entire quantities together ^2 

gives 20 + 13 -i- 7 = 40. Reduce the fractions to ~r| 

Jo j 



higher terms having a common denominator and add, 

. . • • 16 , 12 . 3 31 ,7 tn . , 7 
obtaining -— + srr + "rrr — -srr = 1 



24 



24 



24 24 



24 



40 + X 2l " 



20H 
13i| 

7/t 



41— as the final result. 



40fi 
-40+1^-41 A 



2. Add ab 4- ~ and 2ah + 4- 

Solution. The sum of the entire 
quantities is Sab. The least common denom- 
inator for -7— and -— - is 6, and the new 
2 o 

forms of the fractions become —r- and — — 

6 b 

giving -— * as their sum. As — cannot be 

reduced to lower terms, the result is com- 
plete as shown in the margin. 



ab + — - ab + — 



b 


- 2,6 + » 




«+T 



Find the sum of 



1. I, Ai and i|. 

2. A, tVj and T 6 5 - 

3. 6f, 24, and 4 T V 

4. - 2 -and T - 

* R 1 M 

5. -y and -p- 



PROBLEMS FOR PRACTICE 

Ans. 1 1 



Ans. i3 T Vi 



rn + MN 

Ans. - - vp - 



42 PRACTICAL MATHEMATICS 

i 4<£ . , , d . n 7 d 

6. d + — and 4d + — . Ans. 6d + — 

o J o 

c c 

7. a + ~o" a nd t -f- — . 

8. A room is 32| feet long and 29-J feet wide. What is the 
distance around the room? 

9. Three castings weigh respectively 225|, 232 T 6 ? , and 240J 
pounds. What is their total weight? 

10. A steel rod is to be cut into five pieces; the first to be 4J- 
inches long, the second 3f inches, the third 5-fe inches, the fourth 
4/^ inches, and the fifth lyf inches. Find the length of the rod 
required. 

11. A casting weighing 18f pounds has had 2 J pounds of 
metal removed by the planer. How much did the original casting 
weigh? 

SUBTRACTION OF FRACTIONS 

3 1 . Fractions may be subtracted only when they have a common 
denominator, and express quantities of like units. 

Hence to subtract proper fractions reduce the given fractions to 
their equivalents, having a least common denominator, and write 
the difference of the numerators over the common denominator. 

r o 

Examples. 1. Find the difference between -— ■ and — . 

6 15 

Solution. The fractions, when reduced 
to the least common denominator, become ^ = 

——and -^-r] their difference is, therefore, -— -, tV = i% 

or reduced to lowest terms -— -. 

2. Find the difference between — and — . 

p 2p 

Solution. The fraction - — may be 

V 

reduced to -£-. Then, subtracting, ~~ - 

a 2a la _ la_ a 
2p == 2p~2p~2p° T 2p 

To subtract mixed quantities, subtract the fractional and entire 
parts separately, and add the remainders. If the mixed quantities 
are small, they may be reduced to improper fractions and subtracted. 



9 _ 3 



a 


2a 




V 


= 2p 




2a 


a 


a 


Tp- 


~T P ~ 


Y P 



PRACTICAL MATHEMATICS 43 

Examples. 1. From 27f subtract 14$, 

Solution*. Subtracting the entire quantities, 27 
— 14 = 13. Reduce the fractions to higher terms hav- -'8 := -'!*" 

ing a common denominator am! subtract, obtaining 
20_ 15 _ _5_ 

24 24 24 ' 

2. From 2fJ subtract lj. 

Solution. The common denominator is 14, 

v oil 39 , , 2 9 Is 

hence 2— - — , and 1- = -j- = _. = 

The difference is ^j, which may be reduced to 1' = |J 

the lowest terms by dividing both numerator ami i i~ "° $ = If 

3 
denominator by 7 giving— = 1$ 

To subtract a fraction or mixed quantity from an entire quantity, 
or from a mixed quantity in which the fraction of the minuend is 
less than the fraction of the subtrahend, one unit of the integer in the 
minuend must be written as a fraction. This is shown by the follow- 
ing example. 

9 

3. From 17 take — . 

Solution. First write one of the 
17 units as a fraction having 11 parts. ^ = 16 „ m j nuend 

Thus, 17 - 16 + -ip Now the subtrac- ^_ subtrahend 

tion may be accomplished without 16 rr 

difficulty, giving 16yp 

In the case of letters, subtraction may be made after writing 
the entire quantity, as shown in the margin, 

4. From m take — . 

P 
Solution. The entire quantity m may be written 

in fractional form having the desired denominator by 

multiplying and dividing m by p, thus giving 
The subtraction is then indicated as shown. 

PROBLEMS FOR PRACTICB 

1. From \\ subtract |. 

2. From \ subtract \. Ans. 3 4 . 

3. From f£ subtract |f. Ans. 



m = 


llll) 

V 








in}, 


V 


= 


111 J) — 


a 


I 1 


V 





18 



44 PRACTICAL MATHEMATICS 

4. From ^f subtract T 8 y 

5. From J f- subtract T \- Ans. £-§-, 

6. From jf subtract \\ 

7. From 2| subtract 1| 

8. From | J subtract 3 9 T Ans. ||| 

9. A flagstaff 50 T % feet high was broken off in a storm so that 
43^0 feet remained standing. How much was broken off? 

10. A box and its contents weighed 75f pounds. The box 
alone weighed 3 T 5 e- pounds. What was the weight of its contents? 

MULTIPLICATION OF FRACTIONS 

32. Multiplication of fractions requires no such reduction of 
the fractions as was found necessary in addition and subtraction, 
Multiplying the numerator of a fraction multiplies the number of 
fractional units, their size remaining the same, and dividing the 
denominator multiplies the size of the fractional units, the number 
remaining the same. In many solutions cancellation will be found 
useful. 

• To multiply a fraction by an entire quantity, multiply the 
numerator of the fraction by the whole number, 'and write the prod- 
uct over the denominator, or divide the denominator of the fraction 
by the entire quantity, when it can be done without a remainder, 
and write the quotient as the denominator. Then reduce to lowest 
terms, and if necessary to a mixed quantity. 

7 
Examples. 1. Multiply — by 4. 

8 

Solution a. Multiplying the numer- 
ator by 4 increases the number of frac- 

28 7 2 8 7 

tional units from 7 to 28, giving-^-, which, X 4 - = = 3^- 

8 8 2 

reduced to lower term^, becomes -^- and 

finally as a mixed number 3£. 

Solution b. Dividing the denominator by the ' V 4 = 

divisor is equivalent to cancellation. 4 divided into f> ' 2 

8 is 2. The denominator is then 2, making the frac- 2 

tional unit 4 times as large. By the usual process — - 7 _ n^ 



is changed to 3^. 



2 






PRACTICAL MATHEMATICS 45 

7 

2. Multiply -r- by 3p. 

Solution a. Multiply the numerator 7 7 s/ ;}» 21d 

of the fraction by 3;). and place over the r~~ X •>/> = - 



21s 

denominator tip. — * may be reduced 

by dividing out p in the numerator and 
denominator. 



Op (')[) (')}) 



21p 
6p 


21 3 1 
-- ^ = 3 3 
6 6 2 




1 




2 




2 2 



Solution 6. The same result would have 
been obtained more briefly by dividing the de 
nominator of the fraction by 3/). By cancellation 

the result — is immediately obtained. 

To multiply one fraction by another, multiply together the two 
numerators and place this product over the product of the two de- 
nominators, or indicate the multiplication, and cancel wherever 
possible. 

3. Multiply — by \ . 
8 5 

Solution a. The 2 of the multiplier 

makes the product larger, and therefore, r r> e; v o i n 

the numerator of the multiplicand is mul- x = — = - ■= 

tipliedby2. The 5 of the multiplier makes 8 '5 8X5 40 4 
the product smaller, and hence, the de- 
nominator is multiplied by 5. 

Solution b. The solution in the margin 5 2 1 

shows how the multiplication may be more briefly — - X — - ■=■ — 

performed by cancellation. P 4 



1 



4. Multiply _ by -r-. 



J- v — - d x 2a - 2 °d 
Solution. By the usual method the Ah 4 y / : ~ "TTT 

numerator is equal to the product of d and « / ? 

2a. or 2ad, and the denominator is 46. - — = 

4// 26 

To multiply mixed quantities, change thorn to improper fractions 

and multiply as before. 

5. Multiply 2 " by 5 1 . 

1 • 7 " 4 4 :: 

noN. Write these aa improper Fractions. J ft 21 

-w- and - 1 — is obtained. Cancellation gives 12 at once. 7 K 4 

^ 4 li 



46 PRACTICAL MATHEMATICS 

To multiply mixed quantities and whole numbers or fractions \ 
change the mixed quantities and whole numbers to improper frac- 
tions and multiply. 

2 5 7 
6. Find the product of 2—, 2—, — , and 6p. 

2 2 

Solution: 2§ x * x £ x tf_ M 

PROBLEMS FOR PRACTICE 

Find the product of 

1. 4 and |. Ans. T 5 ¥ 

2. 

3. 

4. 

5. Ilfand8f Ans. w 

6. 

7. 

8 3 . 4 . 10. nnA 1 2 

• "6 > ~9> TS> dna "g"T' 

9. A water-tube boiler has a grate surface of 27 T \ square 
feet. It burns 15 f pounds of coal per square foot per hour. How 
much does it burn in 1^ hours? 

10. What will 142 yards of curbing cost at $6 J per yd.? 

11. What will 17^ tons of coal cost at $4 T 7 o per ton? 

12. A point on the fly wheel of an engine travels 16f feet per 
revolution. How far does it travel in 29| revolutions? How long 
would it take if the wheel made one revolution in 2f seconds? 

DIVISION OF FRACTIONS 

33. Division of fractions is the reverse of multiplication. 
Dividing the numerator of a fraction divides the number of frac- 
tional units, their size remaining the same, and multiplying the 
denominator also divides the size of the fractional units, the number 
remaining the same. This latter process is equivalent to inverting 

7 3 7 4 

the divisor and multiplying. Thus — -*■ — and — X — are 

8 4 8 3 



4 and 4- 


Jul o n H ^L 

9 <*"U 37 • 


17 <lIlu 136* 


12| and 11-J. 


llf and 8f 


12fand7. \ 


6|;|;and4f 
4; 4; If; and 



PRACTICAL MATHEMATICS 47 

equivalent expressions. The fraction after inverting is called the 
reciprocal of the original fraction. 

To divide a fraction by an entire quantity, change the entire 
quantity to a fractional form with one as a denominator; invert the 
fractional form and multiply; or, divide the numerator of the frac- 
tion by the entire quantity. 

Examples. 1. Divide -^- by 4. 
Solution. The reciprocal of 4 is — — ■ n 

This multiplied by-r-, using, cancellation gives _ _:_ 4 = __y_ _ 

2 • 9 9 • 9** 9 

the quotient -5-. 1 

2. Divide — by a. 


1 
Solution. The reciprocal of a is a 0, \ \ 

multiplied by the fraction, and by cancella- ~h~^ aZ= ~h^~ot = ~f 



1 



tion the result is obtained 

3. Divide 13 by y. 

Solution. Invert the 
fraction and multiply. Multi- 3 1Q 7 91 1 

plying 13 by the reciprocal of 13 .+ -=-- 13 X-=- - -«- = 30 -5- 

3 91 1 000 

-=- gives at once -^- or 30—. 

7 00 



d 
4. Divide d by — 



d__fixc 



d -. — - = r . = c 
Solution. The reciprocal of the fraction C fl 

is multiplied by d, and cancellation is used. 1 

To divide a fraction by a fraction, invert the divisor, that is, the 
second fraction, and then proceed as in multiplication of fractions. 

5. Divide — by — -. 

Solution. The multiplication is 

3 2 

accomplished by multiplying — by the 

7 A- i .JL = _lx— = — 

reciprocal of-—, using cancellation. The 4 8^7 7 

result -=- is obtained at once. 



48 PRACTICAL MATHEMATICS 

6. Divide — by — . 

P c 

Solution. The process is exactly i 1 7 

the same as used in the preceding = = X = 

examples. V c p a ap 

When division involves a mixed number, the mixed number may 
be reduced to an improper fraction and one of the preceding methods 
used. 

7. Divide 2 J by 1J. 
Solution. 

1 4 

2-L-1! A„ii> i-vi- A=iJL 

2 ' 8 ' 2 ' 8' ' ?• lf> 3 3 

1 3 
When a fraction takes the form called a compound fraction, that 

i 
is, having a fraction in numerator or denominator or both, as -j 

the division is handled as in the examples just discussed. That 

.7,71 7 7 

is, — -f- 4 = — X 



4 8 X 4 32 



9 

TT 



8. Divide Y by f. 

t> 

9 2 9 5 

Solution. The compound fraction — -f- — = — X — 

is first reduced to a simple fraction; the *■*■ " ■*-■*■ ^ 

division by the second fraction is then / 9 v 5\ , 3_ 9 5 4 

indicated, and this second fraction inverted 111 2/ " 4~~1123 

so as to allow multiplication. Cancella- Q ~ 

tion reduces the fractions to the final 

result. =V- X -X^ = - = 2 S 



11 % 3 11 11 
1 1 , 



Divide : 
1 
2. 
3. 
4. 
5. 



PROBLEMS FOR PRACTICE 



J by 4- An, HI 6 . fl hy l Ans _U 

i by J. ^6 5 

tf by A. m* _&_ 

if by s\. a a 

7fby4|. Ans. Iff 

A railroad 16J miles long cost $66,937; find cost per mile. 

A steam pump delivers 2| gallons per stroke. It delivers 



, 



PRACTICAL MATHEMATICS 49 

330 gallons in 2| minutes. How many strokes docs it make per 
minute? Suggestion: First find gallons delivered per minute. 

DECIMALS 

34. Decimal Fractions. It has been shown that common 

fractions may be reduced to higher or lower terms, /. v., to fractions 

3 
having any desired denominator. For example, the fraction — may 

be reduced to one having a denominator of 12 by multiplying both 

terms by a number which will change the denominator to 12, viz, 

Q ° 

by 3. Thus— -x -jr = —> which is the desired fraction. The 

■x O 1 -j 

same fraction may have the denominators, 20, 32, 40, GO," and 80 by 

multiplying both terms successively by 5, 8, 10, 15, and 20, giving 

. ' . 1 5 2 4 3 4 5 6 . „' ,. . . 

the tractions-—, — , — , — , and 7777 > all of these fractions are 
2 3 2 4 6 8 

qual value with the original fraction. 

Now, suppose it is desirable to reduce all fractions to a standard 

denominator of 10 or any poicer of 10, such as 100, 1,000, etc. Under 



7> 



/•) 



this system the fraction — — will have successivelv the forms — » - 

4 J 10 100 

> etc., while another fraction like -r- will have the forms — > 
10UU o 1 U 

o— 1 ^ — o 

-77777' and -7777777 • Notice that the first form for — - and the first 
1UU 1UUU 4 

and second for — have fractions in the numerators. This makes 

an awkward combination which may always be avoided in deal- 
ing with such fractions by using the higher forms which contain 

P . . , 75 875 

no fraction in the numerator, as -j-7-7- and • 

However, it will be evident that if all fraction- are 50 reduced, 
many of them will have fractional numerators, even though the 
denominators are made higher powers of 10, and hence, in cutting 
off the extra fractions certain errors will be made. It will be 

ied in the section on Per Cent thai such errors, if they are small, 

are usually neglected. The fraction ./for example, may be given 



50 PRACTICAL MATHEMATICS 

TAO 9 

a denominator of 10,000, making a fraction inr JJy which may be 

., , . 769 77 ., 

considered equal to 1onon or even ttjtjtt without any appreci- 
able difference. 

n i * . 75 875 '77 

buch tractions, as t™ j moo ' and • > etc., having powers 

of 10 as denominators are called decimal fractions. In order to 
simplify this system and thus avoid writing the denominator, decimal 
fractions are expressed in another form which consists in writing only 
the numerator and placing a point (.) so that the number of places 
on the right of it shall be equal to the number of ciphers in the 
denominator. The decimal fractions given above, when expressed 
in this way, become .75,* .875, .077, and when so used they are called 
decimals. 

The point (.) is called the decimal point and its office is to mark 
the beginning of the decimal or separate it from a whole number. 

A pure decimal has only decimal places, as, .93678. 

A mixed decimal consists of a whole number and a decimal, 
as 364.23. 

The decimal system is merely an extension of the ordinary num- 
ber system to the right of units place, with a decimal point to indicate 
the boundary line. This is clearly shown in Table V. 

TABLE V 





















J3 






- 












































co 


-d 
G 






CO 

-G 










12 








•5 

•rt 


co 

3 




CO 


G 
O 


-a 












£ 


CO 


G 
cd 

3 


O 


co 


G 
O 


1 


a 

co 

G 






-£2 


"5 
B 


CO 


T3 

^3 


G 

3 


O 

5 


-2 


G 
O 




T3 


O 


g 


g 


o 


a 


G 


O 


G 


(-! 


£3 


G 


G 


,G 


G 


<l; 


a 


a> 


cu 


G 


*G 


CD 


G 




a> 


G 


H 


ffl 


H 


P 


a 


H 


ffl 


H 


H 


M 


H 


w 


6 


4 


3 


9 


• 


3 


6 


5 


7 


9 


8 


2 


3 



In reading the decimals, use the names just given to represent 
the places, omitting the word and except at the decimal point. The 



PRACTICAL MATHEMATICS 51 

following examples give the proper reading and the corresponding 
figures for a few cases : 

1. Nine tenths 0.9 

2. Ninety-five hundredths 0.95 

3. Nine hundred fifty-four thousandths 0.954 

4. Six thousand one ten-thousandths. . 0.0001 

5. Six and one ten-thousandth 6.0001 

The number shown in Table V is read thus: Six thousand, four 

hundred thirty-nine, and thirty-six million, five hundred seventy- 
nine thousand, eight hundred twenty-three hundred-millionths. 

The United States and some other countries use the decimal 
system in money. For instance, take the sum of $13.74. This 
may be considered to be made up of 1 ten dollar bill (tens position), 
3 one dollar bills (units place, dollars being considered units), 7 
dimes or tenths of a dollar (tenths position), and 4 cents or hun- 
dredths of a dollar (hundredths position). It is read thirteen dollars 
and seventy-four cents, and means thirteen dollars and seventy-four 
hundredths of a dollar. It is, of course, better to say seventy-four 
hundredths than seven tenths plus four hundredths, just as it is better 
to say seventy-four cents than to say seven dimes and four cents. 

PROBLEMS FOR PRACTICE 

Write in words: 

1. .965 4. .10792 

2. 3.8506 5. .010952 

3. 5.0061 6. .4563 

Write in figures: 

7. Seven thousand eight hundred forty-nine hundred- 
thousandths. 

8. Sixteen thousandths. 

9. One hundred thirteen ten-thousandths. 

10. Six hundred and thirty-three thousand seven hundred fifty- 
eight millionths. Ans. 600.033758 

11. Twenty-nine hundredths. 

12. Twelve and twenty-seven hundredths. 

13. Four hundred seventy-two and four hundred eighty-seven 
ten-thousandths. 



52 PRACTICAL MATHEMATICS 

ADDITION OF DECIMALS 

35. In addition of decimals, place the numbers so that the 
decimal points will be directly under each other, regardless of the 
number of figures. Having done this, add the figures as in addition 
of whole numbers. The last step is to place a decimal point in the 
sum directly under the column of decimal points. 

Example. Add 1.75; 62.625; and 3.937. 

1.75 

Solution. Following the process given - above and placing nr> ncyr 

the decimal point in the sum directly under the column of decimal o 0,07 

points, 68.312 is obtained. ~ - — * 

68.312 

> The greatest accuracy must be exercised in using decimals. A 
decimal point is more important than any figure, because a mis- 
placed decimal point increases the error at least ten times. 

SUBTRACTION OF DECIMALS 

36. In subtraction of decimals, the process is exactly the same 
as in the case of whole numbers. 

Place the subtrahend under the minuend with the decimal points 
directly under each other. Subtract as in whole numbers. In the 
remainder, place the decimal point in the same column and 
directly under the other decimal points. 

Examples. 1. From 5.17 subtract .01. 



Solution. Subtract as in whole numbers paying careful 
attention to the decimal point. 

2. From 128 subtract 96.307. 



17 
01 



5.16 



Solution. This is an example in which the minuend is a 
whole number, which necessitates placing a decimal point to the 
right of the whole number, and annexing ciphers. In doing this .. ^o r\r\r\ 
the integer is changed to a mixed decimal. Adding ciphers to the _ ' „ _ 

right of the decimal point multiplies both the numerator and de- ' 

nominator of the decimal fraction by the same number, which, 31.693 
although changing its form, does not change its value. Carrying 
out the subtraction as in whole numbers, the remainder 31.693 is 
obtained. 

3. From 134.089 subtract 93. 



Solution. Placing a decimal point at the right of 93, 
subtract as before. 



134.089 
93. 
41.089 



PRACTICAL MATHEMATICS 53 

PROBLEMS FOR PRACTICE 

1. Prom 25. 38 take 14.05 

2. From 39 . 85 take 29 . 755 Ans. 10.00,") 

3. From 72. ISO take 35.976 Ans. 30.213 

4. From 78.896 take 53.5987 

5. From 21.12 take 12.31 

6. From 0.325 take 1.0345 

7. From 0.45 take 2.3375 

S. From SI .35 take 11.679 Am. 00.071 

0. From a cistern that contained 30.5 barrels of water, 25.75 

barrels were drawn off. How much water remained in the cistern? 

10. A hundred pounds of coke were found to contain 5.79 
pounds of ash and .507 pounds of sulphur. The rot was com- 
bined carbon. How much combined carbon was there? 

11. In 1 pound of brazing metal there are .5 pound of copper 
and . 125 pound of tin. The remainder is zinc. How much zinc 
is there? 

12. An iron casting weighed in the rough 22.75 pounds, and 
when finished, it weighed only 16.875 pounds. How much had 
been taken off in the process ? 

MULTIPLICATION OF DECIMALS 

37. In multiplying decimals, proceed as with whole numbers, 
paying no attention to the decimal point until all the figures in the 
product are obtained. Then point off as many places in this product 
as the total number of places in both multiplicand and multiplier, 
prefixing ciphers if necessary. 

Examples. 1. Multiply .397 by 41. 

.397 

Solution. Since there are three decimal places in the 
multiplicand and none in the multiplier, point off 3 + places 397 

in the product. 1588 

16.277 
2. Multiply .027 by .05. 

Solution*. First write the decimals so that t be multiplication .027 

may be most readily performed. 27 is multiplied by 5 as it both Q5 

were integers. Ciphers are then prefixed to 135 until the product 
contains 3+2 = 5 decimal places. 



00135 



54 PRACTICAL MATHEMATICS 

To multiply a decimal or a mixed number expressed as a decimal 
by 10, 100, 1,000, etc., move the decimal point as many places to 
the right as there are ciphers in the multiplier. If there are not 
figures enough for this, annex sufficient ciphers. 

Observe the following: 

.046X1 =.046 .046X100 = 4.6 

.046X10= .46 .046X10000 = 460. 

In the last two multiplications the product contains whole 
numbers. The reason for this is that after the decimal point has 
been moved the required number of places to the right, the cipher 
which comes before is dropped since it has no effect on the value. 
In the last case it is necessary to annex a cipher to the 46 to give the 
number of places required by the four ciphers in the multiplier. 

After a multiplication is completed, if ciphers occur at the 
right of the decimal point with no figures following, the ciphers may 
be dropped. Thus 12.4X10.5 = 130.20. The result should, there- 
fore, be written 130.2. 

PROBLEMS FOR PRACTICE 

Find the product of 

1. .876 and .375 Ans. .3285 

2. 72.2 and .055 

3. 3.62 and .0037 

4. 15.8 and .0855 

5. 2.53 and .00635 Ans. .0160655 

6. .765 and .067 

7. 18.46 and 1.007 Ans. 18.58922 

8. .00076 and .0015 

9. Thirty-four million and twenty-six millionths. 

10. Eight hundred and forty-two thousandths and five hundred 
thousand. 

11. A U. S. gallon of water weighs 8 .335 pounds. What is the 
weight of 17 . 3 gallons? Ans. 144 . 1955 lbs. 

12. A steamship sailed at an average speed of 325.75 miles 
per day. If another steamer sailed from the same port at the same 
time and in the same direction at the rate of 395.25 miles per day, 
how far apart were they in 5.5 days? Ans. 382.25 miles. 



PRACTICAL MATHEMATICS 55 

13. A cubic foot of water weighs 62.3 pounds. How much 
will 177 . 3 cubic feet of water weigh? Aus. 1 1 ,04,") . 70 lbs, 

14. A steam pump delivers 26.4 gallons per stroke. A gallon 
weighs 8.335 pounds. What weight of water will he delivered in 117 

strokes? 

15. In one pound of phosphor bronze, .925 is copper, .07 is tin, 

and .005 is phosphorus. How much of each (copper, tin, and 
phosphorus) is there in 309.5 pounds? 

10. A round bar of rolled iron 2^V inches in diameter weighs 
11.1 pounds per foot. "What is the weight of a bar of the same 
diameter and material which is 9.33 feet long? 

17. A train made an average speed of 1.33 miles per minute. 
How many miles did it cover in 17.5 minutes? 

DIVISION OF DECIMALS 

38. In division of decimals proceed as with whole numbers 
paying no attention to the decimal point until the quotient is ob- 
tained. Then point off in the quotient as many decimal places as 
those in the dividend exceed those in the divisor. This, it will be 
noted, is the reverse of the process just given in multiplication of 
decimals. 

It should be remembered that while the dividend may contain 
more decimal places than the divisor, it must contain at least as many- 
To bring this about, annex as many ciphers as necessary to the 
right of the decimal point in the dividend. 

'When the division does not come out evenly, annex ciphers to 
the dividend and continue the division so as to give at least two 
decimal places in the quotient. 

Examples. 1 . Divide 30 . 744 by 24. 

1.531 



24)30.744 

24 
SOLUTION. No attention is paid to the decimal — — 

point until all the figures of the quotient are obtained. 1-' 

:iere are no decimal places in the divisor and three 120 

in the dividend, the number of decimal places in the 74" 

quotient is three. 72 

li~ 

24 



56 



PRACTICAL MATHEMATICS 



49. 



.002 



2. Divide .196 by .004. 

Solution. There are three decimal places in the 
divisor and three in the dividend. Therefore, none will be .004 ) .196 

found in the quotient. The decimal point is placed to the 
right of the result. 

3. Divide .0027 by 1.35. 

Solution. It is seen that 135 is not contained in 
27. Therefore, annex a cipher to the dividend making 
270, and the divisor 135 is contained twice without a 
remainder. Since a cipher was added to the dividend, 
it contains five decimal places, the divisor contains two, 
and hence, in the quotient there must be five minus two, 
or three decimal places. In order to have three decimal 
places, two ciphers must be prefixed to the quotient, 
and the decimal point placed before them. 

4. Divide 7 by 8. 

Solution. As 8 is not contained in 7 an integral 
number of times it is necessary to annex ciphers, the 

decimal point being placed to the right of 7. The 8)7.000 

division is then carried out as with whole numbers and .875 

the number of decimal places in the quotient is equal to 
the number in the dividend. 



1.35). 00270 

270 





By performing the division which the fraction only expresses, a 

common fraction becomes a decimal. Thus, -, — , — , and — , may 
J 8 24 16 32 J 

be expressed as decimals by dividing 3 by 8, 5 by 24, 7 by 16, and 15 

by 32, giving respectively .375, .208 + *, .437 + , and .469-. 



PROBLEMS FOR PRACTICE 



Divide : 
1. 


183.375 by 489 


Ans. 


.375 


2. 


67.8632 by 32.8 






3. 


67.56785 by .035. 






4. 


.567891 by 8.2 


Ans. 


.06925+ 


5. 


. 1728 by 100 


Ans. 


.001728 


6. 


13.50192 by 1.38 






7. 


783.5 by 6.25 






8. 


983 by 6.6 


Ans. 


148.93+ 



* The + and — signs after the decimals indicate that the true values are slightly 
more or less than the values given. 






PRACTICAL MATHEMATICS 57 

0. 3 by 8 Ans. .375 

10. 1.95 by .45 

11. How much gas at SI. 25 per thousand cubic fvvt ran be 
bought for 817.50? 

12. The distance between two places is 107.75 miles. How 
long will it take a steamer to run the distance if she makes, on the 
average, 12.5 miles per hour? 

13. If a freight train runs at the rate of 15.75 miles per hour, 
how long will it take it to run 1S9 miles? 

14. A lot of 22,840 railroad ties cost $39,867.22. What was 
the cost per hundred? 

15. Seven readings of a dynamometer gave the following horse- 
powers: 17.31, 17.95, 18.13, 17.79, 17.87, 17.63, and 17.47. What 
was the average reading? 

Reduce to decimals: 

17. -1 19. A 

4 16 

20. Change .756 to the nearest 12th. Ans. ,', 

21. Change .564 to the nearest 64th. 

PER CENT 

39. The idea of a fraction of anything, whether it be a common 

fraction like — or a decimal fraction like .001, should now T -be quite 
1 o 

clear. The fraction — — merely serves to show that something has 
been divided into 15 parts and one of these parts has been taken. 
If — of the same thing were taken, this portion would evidently 

be smaller than the — . Again, a given quantity is a bigger part of 
a small group than it is of a large group. The results of one day's 
excavating may be — of all the work to be done on one section of 

a tunnel, but only of the total work if there are 100 sections. 

o< M )0 



58 PRACTICAL MATHEMATICS 

The coin which a man with a bank account gives to a boy in 
exchange for a small service is an insignificant part of the man's 
money, but becomes at once in the boy's hands a large and important 
portion of his capital. 

Evidently, therefore, there must be some standard number by 
which to determine the importance of quantities which are to be 
measured or compared. This standard group number is 100 and all 
changes in numbers can be so reduced as to be expressed in terms 
of 100 parts; when so reduced the result is called the per cent of 
change in that number (written % and meaning by the hundred). 
For example, a change of 2 parts in 200, that is, a change from 200 
to 198, may be reduced by dividing by 2, to equal 1 part in 100, 
or 1%; a change of 10 parts in 500 when divided by 5 is found to be 
the same as 2 parts in 100, or 2%. A man who has one hundred $1 
bills and spends one of them, has decreased his capital by 1 part in 
100 or 1%; if he spends $3, $5, or $15, his capital has been dimin- 
ished by 3%, 5%, and 15%, respectively. Again, a man who buys 
corn at 33c and sells at 35c makes 2c on every 33c, 4c on 66c, or 
6c on 99c, say $1; that is, his gain is 6%. 

Fractions may be expressed in terms of per cent. — may be 

15 

2 
expressed as 1 part in 15 or 6— parts in 100, roughly 7%. This 

o 

means that — of any thing, as the weight of a casting, is about 7% 
15 

of it. The fraction means 1 part in 1000, or .1 part in 100, 

or .1%. Similarly -|-= 12-|-%; ±=33j%; -~=W%, etc. 

Examples. 1. What per cent of error is allowed in a shop if a 
steel shaft 2 inches in diameter must be true to the third decimal 
place? 

Solution. Accuracy to the third decimal place means that the shaft 
can have a diameter of 2.001 inches or 1.999 inches. This is 1 part in 2000, 
.5 part in 1000, or .05 part in 100. The permissible error is, therefore, .05%. 

2. A base line 500 feet long was measured by a party of sur- 
veyors. The total error in this measurement proved to be 2 inches. 
What is the per cent of error? 



PRACTICAL MATHEMATICS 59 

Solution. 500 feel = 500 X 12 inches = 6000 inches. An error of 2 
inches is, therefore, an error oi 2 parts in 0000, or I part in 3000. This is \ of 
a part in 1000 or 3 l in 100 — that is, 3 1 ,, of V c or approximately .03 

3. A contractor figured the cost of a certain piece of work at 
$6750; he added 10% for delays and accidents and 20% for profit. 

What was his profit and what was the amount of his bid? 

SOLUTION. The cost price of the work is $0750. 10', added for accidents 
50 X .10 or $675. _"' . added for proiil will he 2 X 075 = $1350. The 
amount of his bid should he. therefore. s ( i750 -f $675 + $1350 = $8775. 

4. What error is made in using the reducing factor 2.5 centi- 
meters to the inch instead of the actual value 2.540 centimeters? 

Solution. 2.540 —2.5 = .04, the error. This is an error of 4 hundredths 
in 254 hundredths or 4 parts in 254 or about 1 in 04. Increasing 64 by \ = 96, 
approximately 100, and increasing 1 by § — If. Therefore, 1 part in 64 equals 
11 parts in 100 or lf%. The error is If %. 

To put the matter in rule form it is necessary to give names to 
the different quantities as follows: 

The quantity of which the per cent is taken is called the base. 

The number of hundredths or % of the base to be taken is 
called the rate. 

The result obtained by taking the required per cent of the base 
is called the percentage. 

Rules, (a) The product of the rate and the base equals the per- 
centage. 

(b) The percentage divided by the base equals the rate. 

(c) The percentage divided by the rate equals the base. 

(d) To change a number to the per cent form multiply by 100 
and annex the sign %. 

(e) To change a number indicating per cent back to the original 

figures, drop the % sign and divide by 100. For example —=. 25 
and this expressed as a per cent equals .25X100 or 25%. Con- 
verselv^fpj^. 

These rules will help in many cases, but for applications to 
shop work or calculations in general, the more informal method 
given above is strongly recommended. 

Examples. 1. Find the per cent of error made by a machinist 
who took a dimension from a drawing as 4.72 inches and finished 
his piece with a dimension of 4.79 inches. 



60 PRACTICAL MATHEMATICS 

Solution a. The difference between the two dimensions is .07 and as 
there arc approximately 470 hundredths in the dimension, the error is 7 parts 
in 470, about 1 in 65, or 1.5 in 100, i. e., 1.5% error. Ans. 

Solution b. The base is 4.72, the percentage is .07, to find the rate. 
.07 -r- 4.72 = .0148+ or 1.48 + %. 

Note. In solution a, approximate values are taken so as to arrive at 
the conclusion without labored calculation. In this way it is easy to calculate 
% of error mentally with sufficient accuracy. In solution b it is necessary to 
move the decimal point of the quotient two places to the right in order to 
obtain the %. 

2. In a town of 8,000 inhabitants, during an epidemic of typhoid 
fever, there were 114 deaths from the disease. Find the per cent 
of deaths. 

Solution. 114 deaths in every 8,000 equals about 14 in 1,000 or 1.4 
in 100, i. e., 1.4% Ans. 

3. A business firm has a factory stock valued at $28,000. 
At inventory time the firm allows a depreciation of 8%. What is 
the loss? 

Solution a. 8% depreciation means $8 loss on every $100, and as there 
are 280 of these $100, the loss is $8 X 280 = $2240. Ans. 

Solution 6. The base is $28,000, the $2gm x Qg = $2m Ang< 
rate 8%. The percentage will be the loss. 

Note. In such problems as No. 3, solution b is the better method. 

4. The total weight of a freight car when loaded is 148,600 
pounds', and the weight of the empty car is 41,700 pounds; what 
per cent of the entire weight is the weight of the empty car? What 
per cent of the entire weight is the freight carried? 

Solution. 41,700 -f- 148,600 =.28+ or 28 + %. Ans. 
Evidently the weight of the freight is (100 - 28) or 72% of the total weight. 

The method of per cent may be used also in determining how far 
to carry the calculations in various problems. The multiplication 
of the two numbers 4.75 and 65.4 gives 310.65. If it is asked whether 
it is necessary to carry the second decimal place (.05) in this prod- 
uct, this could be determined at once by finding what per cent .05 
is of the whole number. .05 in 310.65 is roughly 5 in 31,000 or 1 

in 6,000, which is — in 100 or — %, a very small error. The slide 
60 60 

rule, by which engineers often make important calculations, does 

not give an accuracy of more than — % to — %. Consequently in all 

J 1 U 



PRACTICAL MATHEMATICS 01 

of the problem calculations, the carrying of the results to four or five 
numerical places (this means the actual figures and does not counl 
the ciphers) is usually sufficient. It is with this probable error 
in view that reports on elections, census, etc., are given in round 
numbers. The population of the U. S. in 1900 may have been 
84,755,643, but for all practical purposes 85 million is sufficient. 
A contractor might figure that a piece of work would be worth 
1 12.743.22 but his bid probably would be 812,750. 

PROBLEMS FOR PRACTICE 

Express in two other ways: 

1. ±,±-,±,* 3. 3%,5%,70%. 5.20%,66J%,125%. 

4 8 2.) 8 

9 o 5 12 1 75 4. — , — , — • 6. .001, -62J, .33J. 
l. ._o, .1-,, -'°- *• 12 ' 10 20 

7 What per cent is $25 of S75? $107.03 of $1,946? $7,804 of 
$11,841? Ans. 33i%; 5J%; 66|%. 

8. A merchant lost $3,000 of his capital and had $21,000 
remaining. What per cent of his capital did he lose? Ans. 121' , 

9. The value of the ratio of the circumference to the diameter 
of a circle (usually designated by the Greek letter tt, pi) is 3.1416, 

2 2 

with an approximate value of — . What per cent of error is made 

in this approximation? Ans. .04+% 

10. A steam engine furnishes 350 horse-power to a dynamo, 
which transforms this into electrical energy with a loss of 8%. Find 
the horse-power supplied by the dynamo. 

11. If gunpowder consists of 15% charcoal, how much charcoal 
is required to make up 250 pounds of gunpowder? 

12. Plaster is made from a mixture of 5 bushels of lime and 7 
bushels of screen sand. What proportion of the mixture is sand? 

13. If S6.C0 is 20% of a man's money, how much has he? 

14. A train of gears transmits 04% of the power supplied at 
one end. If 14.3 H. P. is furnished to the train, how much will 
be delivered to the machinery at the other end? Ans. 0. 1 52 II. P. 

15. A certain 200 II. P. steam engine uses only L8.5< [ of the 
energy of the coal. What would be its horse-power if it turned all 
of the energy into useful work? Ans. L081.08+H.P. 



PRACTICAL MATHEMATICS 

PART III 



DENOMINATE NUMBERS 

40. A denominate number is one in which the unit of value is 
established by law or custom. For example, 7 pounds, 6 feet, 10 
kilograms. 

AYhen a denominate number is composed of units of but one 
denomination, as, for example, 3 gallons, it is called a simple number. 
If it contains units of more than one denomination that are related 
to each other, as 6 feet 10 inches, or 7 pounds 5 ounces, it is called 
a compound number. 

The reduction of denominate numbers is the process of chang- 
ing them from one denomination to another without changing their 
value. The reduction may be from a higher to a lower denomination, 
or from a lower to a higher denomination. 

Note. In a decimal system like the Metric System, the units increase and 
decrease by a uniform scale of 10, but in the English System the scale varies. 

MEASURES 

41. A unit of measure is a standard by which a quantity — such 
as length, area, capacity, or weight — is measured. For example, the 
length of a piece of cloth is ascertained by applying the yard or the 
m eter measure; the capacity of a cask by the use of the gallon or the 
liter measure; the weight of a body by the use of the pound or the 
kilogram, etc. 

There are two systems of measurement which are legalized 
in the United States, the English System and the Metric System. 
The former is in common use in the United States and England, the 
latter in all other countries and in out own governmental depart- 
ments. The metric system is introduced here for general informa- 
tion and for those who hold government positions or who are in foreign 
trade. The student is referred to the Appendix for a complete list of 
tables in the English and Metric systems. 



64 PRACTICAL MATHEMATICS 

MEASURES OF EXTENSION 

42. Extension is that property of a body by virtue of which it 
occupies space and has one or more of the dimensions — length, 
breadth, and thickness. 

A line has a single dimension — length — and its measurement is 
accomplished by linear measure. 

A surface or area has two dimensions — length and breadth — and 
its measurement is accomplished by square measure. 

A solid has three dimensions — length, breadth, and thickness — 
and its volume or capacity is obtained by cubic measure. 

The standard units of extension in the United States are the yard 
and the meter. The yard is 36 inches, the meter 39.37 inches. 

43. Linear Measure. The English measure for length or 
distance, called long measure, makes use of the yard as its funda- 
mental unit, with subdivisions for convenience into feet and inches. 
For instance, a merchant sells cloth by the yard; a person measures 
his height in feet, and the length of his arm in inches. Larger units, 
the rod and the mile, are used when distances to be measured be- 
come so great that the small units are not convenient. A complete 
table of long measure is given in the Appendix. For ordinary cal- 
culations it is sufficient to remember the following: 

12 inches (in.) = 1 foot (ft.) 
3 ft. or 36 in. = 1 yard (yd.) 
5280 ft. = 1 mile (mi.) 

The metric system is founded on the meter as the fundamental unit, 
and as it is a decimal system, the smaller and larger units are all 
decimal divisions or multiples of the meter. For example, the centi- 
meter, which is about 2J times smaller than the inch, is, as its name 
indicates, T | o of a meter; while the kilometer, which is about .6 miles 
is 1,000 meters. Remember the following units: 

10 millimeters (mm.) = 1 centimeter (cm.) 
100 centimeters (cm.) = 1 meter (m.) 
1000 meters (m.) = 1 kilometer (km.) 

It is also useful to remember the following approximate values: 
25 mm. or 2.5 cm. = 1 inch, Fig. 2; 30 cm. = 1 ft.; 1 kilometer = 
.6 mi. 



PRACTICAL MATHEMATICS 



65 



Engineers have adopted the decimal plan in connection with the 
English system by using scales and steel tapes with feet divided 



III!!!!!! IIIIIIIII 111 1 1 1ll 1 11111111! IIIIIIIII lllllllll 111111111 IIIIIMII llll|llll|llll|llll 



8 9 



IIIIIIII! 


miliiii 


iiiiIiiii 


111,1 


IJ 


1, 


iiiiIiiii 


ijili 


if 


mi nil 


I'H 


Ul 


JUilu 


1) 1 1 | 


■ 1 


'I 1 '1 


i 


i 


i 


l'l 1 ! 1 


'1' 


! 


'|'ii| 


M 1 


! i 


iii 



WH\ 



to 



uiti 



ilililililiiililihlilililililtltlililililililililililililiiilil 

Fig. 2. Comparison between Centimeter and Inch Scales. 



Fig. 3. 
A Foot 
Square. 



into tenths. They also use Surveyors' long measure in making land 
surveys. (See Appendix.) 

44. Square Measure. A surface has two dimensions, length 
and breadth. 

The area of a surface is defined as the number of units of surface 
it contains, and is equal to the product of its two dimen- 
sions expressed in the same linear units. The unit of 
surface is a square, which is a plane figure bounded by 
four equal sides and having four right angles, Fig. 3. 
A square, each side of which is one inch in length, is 
called a square inch. Squares formed with sides of 1 
foot, 1 meter, 1 mile, etc., are called respectively, 1 square foot, 1 
square meter, 1 square mile, etc. 

A distinction should be clearly made between the terms square 
foot and foot square or between square mile and 
mile square. If Fig. 3 may be supposed to rep- 
resent a square 1 foot on a side, it may be called 
either 1 sq. ft. or 1 ft. square. On the other 
hand, Fig. 4 measures two feet on a side and 
hence it is 2 ft. square, but, as may readily be 
seen, it has not 2 sq. ft. but 4 sq. ft. of area. 
Therefore a cattle ranch covering an area 3 miles 



Fig. 4. A 2-Foot 
Square. 



square has really 9 square miles of surface. 



66 



PRACTICAL MATHEMATICS 



Square measure, therefore, involves units whose names are the 
same as those used in linear measure with the term square in front 
of each. The English system has an extra unit used in measure- 
ments of land, which is called the acre, equal to the area of a square 
about 209 ft. on a side. 

In the metric system, the square centimeter and square meter — 
the latter being about 20% larger than the English square yard — 
are used for small areas. The larger surfaces are measured in ares 
and hectares, the former being 10 meters square (^V acre) and the 
latter equal to 100 ares or 2\ acres. 

45. Cubic Measure. The volume of any solid is obtained by 

cubic measure. The unit of volume is a cube, Fig. 

5, each edge of which is some unit of length; for 

example, the cubic inch, cubic centimeter, cubic 

foot are common units of volume. 

The volume of a body of rectangular figure is 

equal to the product of its three dimensions, each 

expressed in the same linear unit. 
In the case of an irregular body, however, the volume although 
still expressed in cubic measure, must be measured by displacement 
of water. Liquids are classed as irregular bodies but in the English 
system are measured by a different unit, giving rise to the classifica- 
tion as given in Sec. 46. 

In Sec. 44 it was proved that the 
area of a square surface increases as 
the second power of the side of the 
square; i. e., a surface 2 feet square was 
found to cover 2 X 2 or 4 sq. ft. of area. 
In the same manner, Figs. 5 and 6 show 
that a 2-foot cube has 2 X 2 X 2 or 8 
cubic feet of volume; i. e., the volume of 
a cube increases as the third power of 
the length of the side. 



Fig. 5. A 1-Foot 
Cube. 



<ff?W 



Fig. 6. A 2-Foot Cube. 



Note. The use of the multiplication sign in finding the volume has 
given rise to its use in indicating the dimensions of surfaces and solids. Thus 
a 2 X 12 joist means a joist 2 inches thick and 12 inches wide; or a room 
15' X 12' X 10' means a room 15 feet long, 12 feet wide, and 10 feet high. 



PRACTICAL MATHEMATICS 



67 



The units of cubic measure in the English system are the same 
as those in long measure with the prefix cubic; for example, cubic 
inch, cubic yard. The cord (128 cu. ft.) for wood and the perch (24| 
cu. ft.) for stone or masonry are also used. 

In the metric system, the cubic centimeter and cubic meter are 
common, the latter being 30% larger than the cubic yard and used 
in place of it in measuring earth and rock excavations, as well as in 
measuring timber, stone, etc., where the English cord and perch are 
used. 

MEASURES OF CAPACITY 

46. Capacity signifies the extent of volume or space. In the 
English system a lack of unity exists in the measurement of capacity 

because of the use of several kinds of 

measure. For example, the common 
liquid measure and the apothecaries' 
fluid measure are used for liquids, and 
still another kind called dry measure 
for grains, vegetables, etc. 

This complication is avoided in 
the metric system by having the same 
units for all measurements of capacity. 

47. Liquid Measure. Liquid 

measure is used in measuring liquids 
and in estimating the capacity of cis- 
terns, reservoirs, etc. In the English 
system, the most common liquid 
measure makes use of the gallon, 
barrel, etc. The following units should 
be remembered: 



-Drp Quart 



-. Liter 
Li'01/id Quart 



2 pints (pt.) 
4 quarts 
31 \ gallons 



= 1 quart (qt.) 
= 1 gallon (gal.) 
= 1 barrel (bbl.) 



In the metric system the unit is the 
liter, which is 5% larger than the liquid 
Fig. 7. comparison of Liter, Dry quart and 10% smaller than the dry 

Quart, and Liquid Quart. quart Fig. 7. 



(i actual size) 



68 PRACTICAL MATHEMATICS 

Note the following: 

1000 cubic centimeters (c.c.) — 1 liter (1.) 
100 liters = 1 hectoliter. 

48. Dry Measure. Dry measure is used in measuring dry 
substances such as grain, vegetables, salt, etc. In the English system 
the quart, peck, and bushel are used. In the metric system the 
hectoliter serves the same purpose as the United States bushel and is 
equal to about 3 bushels. 

The common units in dry measure are: 

8 quarts* = 1 peck (pk.) 
4 pecks = 1 bushel (bu.) 

MEASURES OF WEIGHT 

49. Weight is a measure of the force of the earth's attraction for 
a body. 

In the English system several units of weight are used, viz, the 
standard Troy weight, which is used in weighing gold, silver, and 
jewels; the more common avoirdupois weight, used in general trade; 
and the apothecaries' weight, used by druggists and physicians. The 
student is referred to the Appendix for the Troy and apothecaries' 
measure. 

50. Avoirdupois Weight. Avoirdupois weight is used to 
measure the weight of objects in general trade. The most useful 
units are as follows: 

16 ounces (oz.) = 1 pound (lb.) 
2000 pounds = 1 ton (t. or T.) 

The ton just given is the short ton and is more generally used 
than the gross ton which is equal to 2,240 lbs. This latter unit is 
now mainly used in the United States custom house and in weighing 
coarser articles, such as coal at the mines. 

51. Metric Weight. In the metric system the unit of weight 
is the gram which is equal to the weight of 1 cubic centimeter of pure 
water at a temperature of 38° F. The following are the most impor- 
tant units: 

1000 grams (g.) = 1 kilogram (kg.) 
1000 kilograms = 1 metric ton. 
*The dry quart is about 15% larger than the liquid quart, Fig. 7. 



PRACTICAL MATHEMATICS 



69 



30 grams approximately equal one ounce, and one kilogram 
equals 2.2 pounds, hence the gram is used wherever the ounce, penny- 
weight, etc., would be used, while the kilo and half kilo replace the 
pound. The metric ion serves the same purpose as the short or gross 
tons. 

MEASURE OF TIME 

52. Time is measured in the same manner and by the same 
unit throughout the civilized world, the unit being the mean solar 
day. 




Fig. 8. Chart of the Calendar Year. 

This mean solar day is obtained by taking the average of all of 
the days of the year, a day being measured from noon of one day to 
noon of the next day. This process is necessary because the position 
of the earth relative to the sun changes as the year advances, and, 
therefore, no one day can be taken as the true day. Evidently the 
time when it is noon, i. e., the instant the sun is passing the north and 
south line, will be different as we pass from east to west. For example, 
London has her noon five hours earlier than New York. This has 
led to the adoption in the United States of four standard times, the 
Eastern, Central, Western, and Pacific,. each one hour later than the 
preceding. Therefore, when it is noon in Washington it is 11 o'clock 



70 PRACTICAL MATHEMATICS 

in Chicago, 10 o'clock in Denver, and 9 o'clock in San Francisco. 
All cities or towns falling in one of the four regions adopt the standard 
time of that section. The divisions of the calendar year are shown 
graphically in Fig. 8. 

MISCELLANEOUS MEASURES 

53. The English money system has for its principal units the 
pound sterling (£.) equal roughly to $5.00, and the shilling, equal to 
25 cents. The French money system has for its unit the franc, equal 
roughly to 20 cents. The German money system makes use of the 
mark, equal to about 25 cents, for its principal unit. 

A useful table of enumeration is given briefly as follows : 

12 units = 1 dozen (doz.) 
12 dozen = 1 gross (gro.) 

Another table much used by stationers is as follows: 

24 sheets = 1 quire 
20 quires — 1 ream 

The student is referred to the Appendix for the complete tables 
to which references have already be^n made together with others 
which are in more or less common use. It is well to have these 
tables fairly well memorized, and problems are given in the following 
sections on the assumption that the required familiarity has been 
attained. 

REDUCTION OF DENOMINATE NUMBERS 

54. The reduction of denominate numbers in the English system 
is accomplished according to the following rules : 

To change a compound denominate number to a simple number of 
lower denomination : Multiply the integer of the highest denomination 
by the number of units of the next lower denomination in one unit of the 
higher denomination, and add to this product the given number of the 
lower denomination. Proceed into lower terms in this manner until 
the required denomination is reached. 

To change a simple denominate number to a compound number 
of higher denominations: Divide the given number by the number of 
units contained in one unit of the next higher denomination. Set 
aside the remainder; then in the same manner divide the quotient thus 



PRACTICAL MATHEMATICS 71 

obtained, and proceed in this way until the required denomination is 
reached. The last quotient and the several remainders will be the 
result sought. 

Examples. 1. Reduce 6 mi. 11G yds. 24 ft. to feet. 

Solution. Since 1 mile = 1760 yds., 6 mi. = 6 X 1760 yds. 
= 10,560 yds.; and with 116 yds. added this becomes 10,676 yds. 
Since 1 yd. = 3 ft., 10,676 yds. = 32,028 ft., and with 24 ft. added, 
this becomes 32,052 feet, Ans. 

2. Reduce 765 liquid pints to higher denominations. 

Solution*. Dividing 765 pints by 2 gives o^ar 

382 qts. 1 pt. Dividing by 4 to reduce to gallons *) tvO ptS. 

gives 95 gal. 2 qts. 1 pt. Dividing by 31.5 to reduce 4) 382 qts. -f 1 pt. 

to bbl. gives 3 bbl. | gal. 2 qts. 1 pt. As 2 qts. = 31 .5) 95 ga l # + 2 qts. 

\ gal. this added to the -\ gal. already obtained gives q ui 1 — T - T T 

as a final result 3 bbl. 1 gal. 1 pt. 6 DD1, "-" 2 ga1, 

55. Reduction in the metric system is accomplished by moving 
the decimal point. 

Example. Reduce 10,450 millimeters, 276 centimeters, and 600 
meters to kilometers. 

Solution. Reduce 10,450 mm. to meters by moving the decimal 
point three places to the left = 10.45 meters. Similarly reduce 276 
cm. to m. by moving the decimal point two places to the left = 2.76 m. 
600 + 2.76 + 10.45 = 613.21 m. This result may be changed to 
kilometers by moving three points to the left = 0.6132 km. 

"When quantities expressed in the English system are to be 
reduced to their equivalents in the metric system, or the reverse, 
the comparative tables, p. 81, are used. For example, if 34 meters 
should be changed to yards, the table shows that 1 meter equals 1.1 
yards; 34 meters equal 34X1.1 yards or 37.4 yards. 

Examples. 1. Change 172 feet to centimeters. 

Solution-. From the table, 1 foot equals .3 m. 172 feet equals 172 X -3 
or 51.6 m. 51.6 m. equals 51.6 X 100 = 5160 cm. 

2. Change 12.4 pounds to grams. 

Solution. From the table, 1 pound equals .45 kilograms. 12.4 pounds 
equals 12.4 X .45 or 5.58 kilograms. 5.58 kilograms equals 5.58 X 1000 or 
5580 grams. 

PROBLEMS FOR PRACTICE 

1. Change 2560 lbs to ounces; to tons. 

2. Change 42 meters to centimeters; to millimeters. 



72 PRACTICAL MATHEMATICS 

3. Reduce 70 yds. to inches; to feet. 

4. Change 35 ft. to centimeters (approx.). 
Reduce 

5. 18 lbs. 12 oz. to ounces. 9. 3 A. 78 sq.ft.to square feet. 

6. £ 8 12 s to shillings. 10. 5 bbl. 12 gal. to gallons. 

7. 75 m. 86 cm. to centimeters. 11. 2\ T. to pounds. 

8. 8 bu. 4 pks. to pecks. 12. $746.50 to dimes. 
Reduce to higher denominations 

13. 1560 ft. 16. 1640 dry pints. 19. 7562 oz., avoir. 

14. 15,760 mm. 17. 685 doz. 20. 375 pints. 

15. 86,400 sec. 18. 659 shillings. 21. 2500 sq. in. 
Using approximate values, change 

22. 120 yds. to meters. 25. 4750 francs to dollars. 

23. 15 s. to dollars. 26. 65 cm. to inches. 

24. 56 kilos to pounds. 27. 450 liters to pints. 

OPERATIONS WITH DENOMINATE NUMBERS 

56. There are two methods of adding, subtracting, multiplying, 
or dividing denominate numbers. One is to reduce the given numbers 
to the lowest denomination mentioned in the example, then per- 
form the required process and reduce the result to higher denomina- 
tions. The other is to perform the process on the numbers as they 
stand, making the necessary reductions during the operations. 

57. Addition and Subtraction. Examples. 1. Find the sum 
of 3 mi. 182 rd. 4 yds. 2 ft.; 304 rd. 1 ft.; and 5 mi. 76 rd. 4 yds. 2 ft. 
Solution a. 

3 mi. 182 rd. 4 yds. 2 ft. = 15,840 + 3,003 + 12 + 2 = 18,857 ft. 

304 rd. yds. 1 ft. = 5,016 + 1 = 5,017 ft. 

5 mi. 76 rd. 4 yds. 2 ft. = 26,400 + 1,254 + 12 + 2 = 27,668 ft. 

51,542 ft. 

51,542 = 9 mi. 243 rd. 4 yds. ft. 6 in. Ans. 

Solution, b. The sum of the right-hand column jxn, rd. yds. ft. 

= 5 ft. = 1 yd. 2 ft.; write down the 2 ft. and add 1 to 3 ]_g2 4 2 

the yds. column. The sum of the yds. column and 1 n or\A n 1 

carried = 9 yds.; 9-^5£ = 3£ yds. — 1 to be added to r 7fi 4 2 

the rds. column. The sum of the rds. column — 1 car- — — — 

ried = 563 rds. 563 -r 320 = 243 rds. + 1 to be added 9 243 4 \ 



1 


2 


4 








3 
3 


1 
2 


1 
1 



PRACTICAL MATHEMATICS 73 

to the mi. column, making it 9 miles. The result is therefore 9 mi. 243 rds. 
'Sk yds. 2 ft. =9 mi. 243 rds. 4 yds. it . 6 in. 

2. Subtract 3 pks. 1 qt. 1 pt. from 1 bu. 2 pks. 4 qts. 

Solution. Mentally take 1 qt. (2 pts.) from the bu. pks. qts. pts. 
second column and place it in the first; 2 — 1 = 1 pt. 
(4 - 1 borrowed) -1=2 qts. 1 bu (4 pks.) + 2 pks. 
= 6; 6—3 = 3 pks. The result is therefore 3 pks. 2 qts. 
lpt. 

PROBLEMS FOR PRACTICE 

Find the sum of 

1. 10 yds. 2 ft. 10 in.; 15 yds. 1 ft. 9 in.; 8 yds. 2 ft. 7 in.; 
18 yds. 1 ft. 11 in.; 16 yds. 2 ft. 8 in. Ans. 12 rd. 4 yds. 2 ft. 9 in. 

2. 12 A. 35 sq. rd.; 14 A. 110 sq. rd.; 15 A. 132 sq. rd.; 11 A. 
96 sq. rd.; 25 A. 100 sq. rd. Ans. 79 A. 153 sq. rd. 

3. 5 t. 6 cwt. 14 lbs, 10 oz.; 7 t. 15 cwt. 36 lbs. 15 oz.; 17 t. 
5 cwt. 84 lbs. 12 oz.; 70 t. 9 cwt. 94 lbs. 11 oz. 

Ans. 100 t. 17 cwt. 31 lbs. 
From 

4. 12 gal. 2 qts. 1 pt. 2 gi. take 6 gal. 3 qts. 1 pt. 3 gi. 

Ans. 5 gal. 2 qts. 1 pt. 3 gi. 

5. 15 yds. 2 ft. 7 in. take 4 yds. 2 ft. 10 in. 

Ans. 10 yds. 2 ft. 9 in. 

6. 25 t. 8 cwt. 75 lbs. 10 oz. take 10 t. 11 cwt. 35 lbs. 15 oz. 

Ans. 14 t. 17 cwt. 39 lbs. 11 oz. 
58. Multiplication and Division. Examples. 1. Multiply 14 
gals. 3 qts. 1 pt. by 7. 
Solution a. 
14 gal. 3 qts. 1 pt. = 112 + 6 + 1 = 119 pts. 
119 pts. X 7 = 833 pts. = 416 qts. 1 pt. = 104 gal. 1 pt. Ans. 

Solution b. (14 gal. 3 qt. 1 pt.) X 7 = 98 gal. 14 gal. 3 qts. 1 pt. 
21 qts. 7 pts. = 98 gal. 24 qts. 1 pt. = 104 gal. qts. 7 

l &' ***' 104 gal. qts. 1 pt. 

2. Divide 14 gal. 3 qts. 1 pt. by 4. 
Solution a. 

14 gal. 3 qts. 1 pt. == 119 pts. 
4 )119 pts. 
29 J pts. = 14 qts. If pts. = 3 gal. 2 qts. If pts. Ans. 



74 PRACTICAL MATHEMATICS 

PROBLEMS FOR PRACTICE 

Multiply 

1. 3 hrs. 20 min. 35 sec. by 5. Ans. 16 hrs. 42 min. 55 sec. 

2. 2 t. 5 cwt. 48 lbs. 15 oz. by 8. Ans. 18 t. 3 cwt. 91 lbs. 8 oz. 

3. 12 cu. yds. 15 cu. ft. by 6. Ans. 75 cu. yds. 9 cu. ft. 
Divide 

4. 15 bu. 3 pks. 5 qts. by 4. Ans. 3 bu. 3 pks. 7 qts. \ pt. 

5. 23 cwt. 68 lbs. 10 oz. by 5. Ans. 4 cwt. 73 lbs. llf oz. 

6. 15 rd. 4 yds. 2 ft. 8 in. by 5. Ans. 3 rd. 2 ft. 11 J in. 

POWERS AND ROOTS 

59. Powers. A power of a quantity is the product of factors, 
each of which is equal to that number. This quantity may be 
simply a number, as 1, 2, 3, or 4, or it may be any letter, as 
a, b, c, or d, which may have any numerical value whatso- 
ever. 

In order to show how many times the quantity is to be used 
as a factor, a small figure is placed to the right and a little above the 
quantity; as 3 4 or a 4 . This small number is called an exponent, 
and shows to what power the quantity is to be raised. Thus, 3 4 
means the fourth power of 3, or 3 X 3 X 3 X 3 = 81; i. e., 3 is 
taken 4 times as a factor. If the letter a is substituted for the figure 
3 and this letter is raised to the fourth power, the result is a X a X a X 
a or a 4 . Since no definite value for a has been given, the result of 
raising the letter to the power can only be indicated ; thus, a 4 . 

The second power of the quantity is called its square. For 
example, 4 is the square of 2, for 2 X 2 = 4; again, a 2 is the square 
of a, for a X a = a 2 . The third power of a quantity is called its 
cube; thus, 8 is the cube of 2, for 2 X 2 X 2 = 8; or a 3 is the cube of a. 

Suppose now it is required to find the square of a 2 , a 2 X a 2 = a 4 , 
for a 2 X a 2 = (a X a) X {a X a) = a 4 ; or, in other words, when 
two like quantities are multiplied together, the exponent of the product 
is equal to the sum of the exponents of the quantities. 

The power of a fraction is obtained by multiplying the numerator 

by itself and the denominator by itself the required number of times. 

2 2X24 

Thus, the second power of — - = 7 = -— ; the third power of 

r 4 4 X 4 16 r 

J_ 1X1X1 = J_ 

7 7X7X7 343' 



PRACTICAL MATHEMATICS 



75 



Examples. 1. What is the cube, or third power of 21? 

21 3 



Solution. Here it is readily seen that raising 
21 to the third power is the same as using 21 three 
times as a factor. 



= 21 X 21 X 21 

21 

21 

21 

42_ 

441 

_21 

441 

882 



9261 



Ans. 



2. What is the cube, or third power of 2a? 

(2a)' 

Solution. Note here, that when raising a 
quantity like 2a to a power, the coefficient 2 and 
the letter a are each raised separately to the required 
power. 



= 2a X 2a X 2a. 
2a 
2a 
4a 2 
2a_ 
8a 3 Ans. 



3. What is the cube of .71? 



Solution. When raising a decimal like 
.71 to a power, care should be taken to correctly 
place the decimal point. In the answer to this 
problem there will be six places to be pointed 
off, thus bringing the decimal point before the 3. 



.71 s = . 



71 X .71 X .71 
.71 

21 

71 

497 



.5041 

.71 

5041 

35287 
.357911 Ans. 



4. What is the fourth power of — ? 



x l X -X-= 3X3X3X3 
5 5 5 "5X5X5X5 



81 
625 



= t^i.. Ans. 



PROBLEMS FOR PRACTICE 

Raise the following quantities to the power indicated: 

1. 47 to the second Ans. 

2. 6m " " second 



2,209 



76 PRACTICAL MATHEMATICS 



3. 


71 to 


the second 


4. 


2.61 " 


' second 


5. 


.13 " 


" third 


6. 


c 2 " 


" third 


7. 


r 


" fifth 


8. 


k " 


" fourth 


9. 


Qk " 


" fourth 



Ans. 5,041 
Ans. 6.8121 



Ans. 1296& 4 

60. Roots. A root of a quantity is one of the equal factors 
which, when multiplied together, give the quantity. Thus, if a cer- 
tain quantity is used twice in order to produce another quantity, then 
the quantity first mentioned is the square root of the second. Thus, 
2 is the square root of 4, for 2 X 2 = 4. If the first quantity must be 
used three times as a factor, it is the cube root of the second quantity; 
thus, the cube root of 8 is 2, for 2 X 2 X 2 = 8; the cube root of 
27 is 3, for 3 X 3 X 3 = 27. This also applies to any root as the 
4th, 5th, etc. Thus, the 5th root of a 5 = a, for a X a X a X a X a 
= a 5 . 

This process of finding the root is merely the reverse of finding 
the power, and is termed extracting the root. 

The radical sign <J when placed before a quantity shows that 

some root of it is to be taken. The root is indicated by a small figure 

3 ; 

called the index placed above the radical sign; thus, -%/ . When 

no index is written, the square root is always understood. 

The following examples show the meaning of the sign and 
index : 

^81 = 9, for 9 X 9 = 81. 

^27 = 3, for 3 X 3 X 3 = 27. 

^81 = 3, for 3 X 3 X 3 X 3 = 81. 

^32 = 2, for 2 X 2 X 2 X 2 X 2 = 32. 

^V~= b, for 6 X b = b\ 

^8b*= 26, for 2b X 2b X 2b = 86 3 . 

^J32b 10 = 2b\ for 2b 2 X 2b 2 X 2b 2 X 2b 2 X 2b 2 = 326 10 . 



PRACTICAL MATHEMATICS 



77 



61. Square Root.* In the above illustrations the required root 
may be readily determined by inspection, but in most cases, this 
method cannot be followed. When the root desired is the square 
root the method illustrated below should be used. 

Examples. 1. Find the square root of 185,761. 

Solution. First point off the num- 
ber into periods of two figures each com- 
mencing at the right of the number. Then 
find the largest number whose square is equal 
to or less than 18. This is foimd to be 4. Set 
down the square of 4 or 16 under 18, and 
place 4 as the first figure of the root. 

Subtract 16 from 18 and to the re- 
mainder annex the next period, obtaining 
257 as the new dividend. The new divisor 
is twice the root already found with a cipher 
annexed, that is 2 X 40 = 80. By trial 80 
is found to be contained 3 times in 257, so 
3 is placed as the second figure of the root. 
Add 3 to SO and multiply this sum 83 by 3. 
Place this product, 249, under 257 and 
subtract. Annex to the remainder the last 
period obtaining 861 as the new dividend. 

The next divisor is twice 430 or 860. 
This is contained in 861 once. Place the 
1 as the third figure of the root, add 1 to 
860, thus making the complete divisor 861. 
As there is no remainder, the number 185761 
is a perfect power, and its square root is 431. 



2 X40 



30 
3 



18 
1G 



57 61 



83)257 
2 X 430 = 860 

1 249 



861)861 
861 



Find the square root of 18,763.8910 



Solution. Point off the periods to 
the left and right of the decimal point, ob- 
taining for the whole number the periods, 1, 
87, and 63^ and for the decimal the periods, 
89 and 10. In the solution given in the 
margin only the complete divisor is shown. 
If this process is not clear, refer to the solu- 
tion given in Example 1. The root must 
have three numerical places to the left of 
the decimal point to correspond to the three 
periods in the whole number; hence, the final 
value of the root is 136.98. 



1 36.98 +__ 
2X10=20 1 87 63.89 10 

Li L_ 

23)87 
2X130 = 260 

6 69 



2X1360 



2X13690 



266)1863 
2720 
9 1596 



2729)26789 
27380 

8 24561 



27388)222810 
219104 



* To extract cube root and roots of higher powers, the use of logarithms is recommended, 
aa explained p. 132, Part IV. 



78 PRACTICAL MATHEMATICS 

3. Find the square root of .001225. 

Solution. In finding the square root of a decimal nQ _ 

like the above, it is seen that the first period contains 

only zeros; consequently the first figure of the root will -00 12 25 

be a zero. Begin by finding the square next smaller 9 

than 12, and proceed as in previous examples. Point 65Y325 

off three places in the root to correspond to the three qoc 

periods in the decimal. 

4. Find the square root of 3 to three decimal places. 

J..732 + 

3.00 00 00 
1_ 

Solution. Add ciphers at the right of the 27)200 

decimal point and separate into periods as usual. 189 

343)7100 
1029 

3462)7100 

Rules. (a) Point off the given number into periods of two figures, 
beginning at the decimal point. If the number contains a whole num- 
ber and a decimal, point off periods to the left of the decimal point 
for the whole number, and to the right for the decimal, annexing, a 
cipher to the last figure of the decimal if necessary. 

(b) Find the greatest number tvhose square is contained in the left- 
hand period. This figure will be the first figure in the root. Subtract 
the square of this number from the left-hand period and annex the 
next period to the remainder to form the new dividend. 

(c) After annexing a cipher to that part of the root already found 
and doubling the result, place the quantity thus obtained on the left for a 
trial divisor. Ascertain how many times this divisor is contained in the 
new dividend, and write the quotient as the next figure of the root. Then 
add the number just placed in the root to the trial divisor, and multiply 
the completed divisor by the last figure in the root. Subtract this prod- 
uct from the dividend, and to the remainder annex the next period for 
the new dividend. Continue the operation until all the periods are used. 

(d) To place the decimal point in the result, if the number is a 
decimal number, point off in the root as many figures from the left as there 
are periods from the left to the decimal point in the original number. 

(e) If after the root is found there is a remainder, it should be 
dropped and a -f sign placed after the root. 



PRACTICAL MATHEMATICS 79 

PROBLEMS FOR PRACTICE 

1. Find the square root of 304,930 Ans. 604.09+ 

2. Find the square root of 825,487.69 Ans. 908.56+ 

3. Find the square root of .003804 Ans. .0621+ 

4. Find the square root of .0503 Ans. .2372+ 

5. Find the square root of 1,873 

6. Find the square root of 0,432 

62. Square Root of Fractions. When the square root of a 
common fraction is desired, first see if the numerator and denominator 
have perfect square roots. If so, write the root of the numerator 
over the root of the denominator. If this is not the case, reduce the 
common fraction to decimal form and find the required root as before. 
If the quantity is a mixed quantity change to decimal form before 
extracting the root. 

The following solutions will illustrate these principles: 

i c 
Examples. 1. Find the square root of — 



(l6 = ^16 = _i 

^49 i/49 ' 

4a 2 

2. Find the square root of — - 

4 166 2 

V~4a? 2a 

V 7 !^ 2 ~ 4b 



Ans. 



Ans. 



169 
3. Find the square root of — - to three decimal places. 

.822+ 

250) 169.000 .67 60 00 

.676 64 

162)360 
324 



1642)3600 
32S4 



Hence, yj— =l/.676 = .822+ Ans, 



80 PRACTICAL MATHEMATICS 

64k 2 



4. Find the square root of 



81m 2 
l/64k 2 8k^ 
l/81m 2 9m 



Ans. 



3 
5. Find the square root of — to two decimal places. 

8 

.37 50 

8 )3.000 36 

.375 121)150 

121 
29 

Hence, yj— = l/^75 = .61 + Ans. 

PROBLEMS FOR PRACTICE 

Find the square root of 

1. ^ 4. | An, .79 + 

2 - ilk ^ - 041+ 5 - I 

3. I 4 6. 8U ° 



81 1216 1 

RATIO AND PROPORTION 

63. Ratio. Ratio is the relation which one quantity bears to 
another quantity of the same kind; or, in other words, it is the quotient 
obtained by dividing the one by the other. The ratio of 15 to 3 is 

5, because 15 contains 3, five times. The ratio of 3 to 15 is — -, because 

5 

3 is — of 15. 
5 

The two quantities given are called the terms of the ratio. The 
first term is called the antecedent; the succeeding one, the consequent. 
The antecedent is the dividend, the consequent is the divisor, and the 
value of the ratio is the quotient resulting from the division. The 
symbol used to express a ratio is two dots (:), which is merely an 



PRACTICAL MATHEMATICS 81 

abbreviated form of the sign of division (-5-) For example, the 
ratios 7 to 11, and a to 6, are written thus, 7:11, and a : 6. The 

same ratios may also be expressed in fractional form thus, — or — . 

^ 11 6 

A ratio like the above is called a direct ratio; that is, one in which 

the antecedent is divided by the consequent. An inverse ratio is a 

direct ratio inverted Thus, if — and — are the direct ratios, the 

11 6 

inverse ratios of the same quantities are -— and — . 

^ 7 a 

As a ratio is always an abstract number, it is possible to have a 

ratio between similar quantities only. For example, it is possible 

to establish a ratio between 75 rivets and 16 rivets, but no ratio 

between 75 rivets and 10 horses. If the ratio — is given, and a 

represents 5 horses and b, 10 horses, it can be seen that, although this 
ratio has been expressed in letters, each letter is a quantity of the 
same kind. 

The value of a ratio is not changed by either dividing or mul- 
tiplying both terms of the ratio by the same number. Thus, for 
example, if 16 : 32 is divided by 2, the ratio becomes 8 : 16; likewise, 
each term may be multiplied by 2, obtaining 32 : 64. From the above 
it is seen that the values of the ratios have not been changed for 

— = — X — = — -r- -x- . Consider the ratio 5a : 106. Divid- 
32 16 2 64 2 

ing each term by 5, the ratio a : 2b is obtained. Again multiplying 

each term of the first ratio by 2, it becomes 10a : 206, which has the 

same value as before because -— - X — - = 7^7-. 

26 10 206 

PROBLEMS FOR PRACTICE 

What is the value of each of the following? 

1. 51 : 17 Ans. 3 



2. 20c : 40a 7 Ans. 

3. 16 : 48 

4. 6k : 8m Ans. 



c 
2d 

3& 
4m 



82 PRACTICAL MATHEMATICS 

SI 

5. What is the inverse ratio of 9 : 81? Ans. — = 9 

9 

6. What is the inverse ratio of a : bf 

7. What is the inverse ratio of 10k : 20r? 

64. Proportion, (a) A proportion is an expression of equality 
between ratios and is indicated by the sign :: or by the equality 
sign =. For example, the proportion 3 : 9 :: 6 : 18 is read 3 is to 
9 as 6 is to 18. The same proportion may be stated thus: 3 over 

9 equals 6 over 18; that is, — == — . Any proportion whatever exist- 

9 18 

ing between letters is expressed in a similar manner ; thus, a:b :: c : d 

. a c 

may also be written — = — • 
b a 

(b) The terms of a proportion are made up of the antecedents and 

a c 
consequents of two ratios. Thus, in the proportion -— = -; > 

b a 

a, b, c, d, are called the terms and are numbered — first, second, third, 

and fourth. This may be illustrated as follows: 

First Second Third Fourth 

3 : 9 : : 6 : 18 

or a : b : : c : d 

The first and fourth terms are called the extremes; the second and 
third terms, the means; thus, 3, 18, and a, d, are the extremes, while 
9, 6, and b, c are the means. 

(c) In all proportions the product of the means is equal to the prod- 
uct of the extremes. Thus, in the proportion 3 : 9= (\: 18, the product 
of the means is 9 X 6 = 54 and the product of the extremes is 3 X 18 

= 54. If the proportion is expressed in fractional form, that is, 

q a 
— = — , it is seen that the law iust stated means that the product of 

9 18 

the numerator of the first fraction with the denominator of the second 
is equal to the product of the numerator of the second fraction with 

3 6 

the denominator of the first. Therefore, in the proportion — = — » 

9 18 
a c 
3X18=9X6 and, in the proportion — = — , ad— be. When 

b d 
taken by themselves the letters, a, b, c, d, may have any values 



PRACTICAL MATHEMATICS S3 

whatsoever, but as soon as they are placed in the proportion, their 

relation to each other becomes fixed. If the proportion has an 

a c 
unknown term to be found, this latter form —= — is perhaps the 

b a 

simpler form in which to arrange the ratios for finding the missing 

term. 

Rules, (d) The product of the means divided by either extreme 
gives the other extreme; or the product of the numerator of the second frac- 
tion with the denominator of the first fraction divided by the numerator of 
the first fraction gives the denominator of the second fraction. 

(e) The product of the extremes divided by either mean gives the 
other mean; or the product of the numerator of the first fraction with 
the denominator of the second fraction divided by the denominator 
of the first fraction gives the numerator of the second fraction. 

In stating a proportion in which one term is unknown, let x 
represent the unknown term and solve for x. 

Examples. 1. Solve for x in the proportion, 18 : 6 = 9 : x, 

18 9 

or - — = — . 
6 x 

Solution. Product of extremes, 18 X x = 9 X 6, product of means 

54 

18 



54 
Dividing each side by 18 x = — = 3 Ans. 



Hence the missing term is 3. 

2. Solve the proportion, 6 : 18 = x : 3. 
Solution. 18 X ^ = 6 X 3 x = 1 Ans. 

3. Solve the proportion, 9 : 3 = 18 : x 

18 X 3 a A 
Solutiox. x = — — = o Ans. 

y 

4. Solve the proportion, 3 : 9 = 6 : x 

6 X Q 1Q A 
Solution. x = — - — = 18 Ans. 
o 

5. Solve for x in the proportion a :b : :c : x 

Solution. This may be written in this form, ax = be, from 

be ax == be 
which x= — . In this case the answer can only be indicated, but 

a be 

if o, b, and c, had the respective values, 5, 6, and 10, their sub- X = — 

fi v 1 O 
stitution would give toxa value of 12, since x = — ^ — - = 12 

5 



84 PRACTICAL MATHEMATICS 

Find the value of x in the following proportions: 
6. k : m : : x : n. 



k x 

Solution. — = — 
m n 




mx = kn 




m 


Ans. 


7. s : r : : x : p. 












s X 

Solution. — = — 

r p 

8. e :/: : .r : A. 




rx = sp 




sp 
x = ~- 
r 


Ans. 


Solution. — = — 
J ' l 




jx — eh 




eh 


Ans. 


PROBLEMS FOR PRACTICE 

Solve for the unknown value of x in the following proportions: 


1 6 : 3 : : m : x 
2. 8 : 9 : : x : 9 


Ans. 


m 
T 


6. 

7. 


21m :3 
a; : 4 : : 


: : x : 7 

3r : s Ans. — 


3. 9 : 81 : : 2 : x 






8. 


384 : 64 


: : x : m 


4. 4 : x : : 10 : 5 






9. 


480 : x 


: : 10 : 48 


5. r : s : :2 : x 






10. 


90 :x : 


: 9 : k Ans. 10& 



65. Any root of both sides of a proportion may be extracted, 
or any power of both sides may be taken, without destroying the 
proportion. 

Examples. 1. Extract the square root of both sides of the 
following proportion : 1 : 9 = 49 :441. 

Solution. Extracting the square root of each term 
1:3 = 7 :21 

1_ 7_ 

3 21 

which equation, by clearing of fractions, shows that the equality of 
the proportion has not been destroyed. 

2. Extract the square root of both sides of the following pro- 
oortion : a 2 :b 2 = c 2 : d 2 . 

Solution. This proportion may be stated thus: 



Extracting the square root 7- = -j 



PRACTICAL MATHEMATICS 

3. Find the cube of the proportion, 1:2 = 2:4 

1 2 
Solution. Changing its form — = — 

Cubing each term — ——- 
8 64 

Thus it is seen that the proportion is not destroyed. 

4. Cube the proportion, -y- = -p 

a 3 c 6 
Solution. Cubing each term t~ = — ^ 

b 3 a 

Thus it is seen the rules are true for both letters and numbers. 

66. Two numbers or quantities are directly proportional when 
they increase or decrease together; in which case their ratio is always 
the same. 

Two numbers are inversely proportional when one increases as 
the other decreases; in which case their product is always the 
same. 

For example, if 8 men each perform 10 units of work per hour, 
they will perform 40 units in 4 hours and 100 units in 10 hours; that 
is, as the time increases the amount of work done increases. This 
is a direct ratio. Xow, on the other hand, if there are say 800 units 
of work to be done on a job, 8 men will finish this work in 10 hours, 
16 men will finish it in 5 hours, 40 men in 2 hours. Here the time 
varies inversely, or oppositely, as the number of men employed, and 
such a ratio is called an inverse ratio. 

67. The following rule will solve any problem in simple pro- 
portion whether it be direct or inverse: 

Rule. Place for the third term of the proportion the number ichose 
units are like the answer sought (80 rivets in Example 1 below; 7 
hours in Examp'e 2). From the conditions of the problem determine 
if the answer (x) is to be greater or less than the third term; if greater, 
place the larger of the two remaining numbers for the second term; if 
less, place it for the first term. Then solve according to the rules given 
for finding the missing term. 

Examples. 1. If a joint 16 feet long requires 80 rivets, what 
number will a joint 11 feet long require? 



86 PRACTICAL MATHEMATICS 

Solution. Let x equal the number of rivets required. 
Then 16 : 11 = SO : x 

16 80 

or — = — 

11 x 

80X11 

x = 55 rivets. Ans. 

The result shows that the shorter joint requires fewer rivets; hence, 
this is called a direct proportion. 

2. A train running 27 miles per hour covers a certain distance 
in 7 hours. How long does it take a train running at 32 miles per 
hour to cover the same distance? 

Solution. It is evident that the train 32 : 27 : : 7 : a; 

running 32 miles per hour will require less qo _ 97 y 7 

time to cover the same distance than the 

train running 27 miles per hour. Thus, the _ ^°^ 

faster train requires less time, and the slower 32 

train more time; or, mother words, the speed x _ 5 g 1 j^g ^ ng 

varies inversely as the time. 

It can readily be seen that the proportion 32 : 27 : : 7 : re is an inverse one. 
If the speed varied as the time, the proportion would be direct and stated as 
follows: 32: 27 : : x : 7. It will be noted in the direct form of this proportion 
just given, that the second and fourth terms refer to the first train and its time, 
while the first and third terms refer to the other train and the unknown time. 
In the inverse proportion the time of the first train is found as the third term. 

PROBLEMS FOR PRACTICE 

1. If the earth moves through 360 degrees in 365| days, how 
far will it move in a lunar month of 29| days? Ans. 29.07+ 

2. How long will it take a gang of 50 workmen to erect the 
walls of an 8 story building, if it requires 4 days for 100 workmen to 
erect one story? 

3. If 15 men can build a wall 12 feet high in one week, how 
many men will it require to raise it 20 feet in the same time? 

4. If it requires 18 hours to saw 10,000 feet of lumber, using 
a 20 h.p. engine, what horsepower will be required to saw the same 
amount in 10 hours? 

68. Negative Quantities. If an ordinary thermometer is con- 
sulted it will be found that the scale divisions have opposite them 
numbers which increase both upwards and downwards from 0. 



PRACTICAL MATHEMATICS 87 

All values below the zero point are considered negative and all values 
above are considered positive; thus a temperature of —6 degrees 
means that the mercury reads (3 degrees below zero; similarly 6 
degrees means that the reading is degrees above zero. 

The boiling point of water on the Fahrenheit thermometer scale 
is 212 degrees. On the other hand, the temperature of liquid air, 
which is a very cold body, is —292 degrees Fahrenheit. Thus it is 
seen that the point is merely a point of reference. In the same 
manner, the of the numerical system is considered as a figure of no 
value and all numbers with the + sign have positive values and all 
numbers with the — sign have negative values. 

Suppose a man's money in the bank is 300 dollars and his debts 
amount to 350 dollars; here his money in the bank may be considered 
as positive and his debts as negative. Thus + 300 — 350 = — 50 
dollars. This —50 shows that the man is in debt 50 dollars, or has 
virtually 50 dollars less than nothing. 

Again when a ship sails a miles to the north of the equator, it is 
sailing in a positive direction as compared with a course to the south 
of the equator. For example, if a ship travels 200 miles north it is 
said to have traversed + 200 miles; on the other hand if it turns about 
and sails in the reverse direction 250 miles, it then will have a position 
50 miles to the south of the starting point, i.e., its position will be — 50 
miles. Expressing this in equation form gives 200 — 250= —50. 

Examples. 1. What is the difference in longitude between two 
places where the longitudes are —80° and +30°? 

Solution. Since one position is 80° 
from the starting point in one direction and 

the other 30° in the opposite direction from . 

the starting point, the difference between oU+«3U = 110 Ans. 

the two places will be represented by the 
equation: 

2. A man has bills receivable to the amount of 500 dollars, 
and bills payable to the amount of 1,000 dollars; how much is he 

worthr 

Solution. Having bills receivable to 
the amount rf $500, tins is what he is worth; S50 0-$1000 = -$500 Ans. 
but having bills payable to the amount of . ' m 

$1,000, this is what he owes. He is actually or ne 1S m debt $500 
worth then: 



88 PRACTICAL MATHEMATICS 

PROBLEMS FOR PRACTICE 

1. The temperature at 6 P. M. is + 14° and during the evening 
it grows colder at the rate of 4° an hour. Required the temperature 
at 9 P. M., at 10 p. M., and at midnight. 

2. What is the difference in latitude between two places where 
the latitudes are +86° and -14°? 

69. Parenthesis. The parenthesis has already been shown to 
indicate that the terms enclosed are to be considered as one quan- 
tity. The following rules indicate the proper use of the parenthesis. 

Rules, (a) When a parenthesis is preceded by the + sign the 
parenthesis may be removed without making any change in the expres- 
sion loithin the parenthesis. 

(b) When a parenthesis is preceded by the — sign the parenthesis 
may be removedifthe sign of every termwithin the parenthesis be changed. 

(c) When a number or letter immediately precedes or folloivs the 
parenthesis, with no sign between, the multiplication sign is understood. 

Examples. 1. Remove the parenthesis from the expression: 
8 + (2 + 6). 

8 + (8) = 8 + 8 = 16 

In this problem, 2 is added to 6 before the parenthesis is re- 
moved, and the result is added to 8 to give the total sum of 16. It 
should be noted that the parenthesis has been removed without 
changing the signs within the parenthesis. 

2. Remove the parenthesis from a + (b + c — d). 

a + (6 + c — d) = a -\- b -\- c — d 

3. Remove the parenthesis from 12 — (6 — 4) 

According to the rule, since the quantity (6 — 4) is preceded by 
a minus sign the removal of the parenthesis makes it necessary to 
change the + 6 to — 6, and the — 4 to + 4. The expression then 
becomes 12 — 6 + 4, or 10. 

If the operation indicated had been performed before removing 
the parenthesis and the result subtracted from 12 the remainder 
would have been the same. 

4. Remove the parentheses from c + k — (m — 1) + (a + b). 
The signs of m and 1 are changed to — and + respectively on the 

removal of the parentheses and the final expression becomes c + k — 
m + 1 + a + b. 



PRACTICAL MATHEMATICS 89 

5. Remove the parentheses from — (c — k + 1 — m + c 2 ) 
+ S - (6 - 8). 

Removing the parentheses and changing' the necessary signs the 
expression becomes — c + A* — 1 + m — c 2 + S — 6 + 8 = — c+k 
- c 2 + m + 9. Ans. 

PROBLEMS FOR PRACTICE 

Remove the parentheses in the following problems: 

1. - (4 + 10) 

2. a + b - (c - ax + 5) 

3. 36 - (14 - 8) + (3 - 10) 

4. c - (a + b) + (c 2 - m) 

EQUATIONS 

70. An equation is an expression of equality between two 

a x 
quantities. It has been proved that the proportion -7- =-7- can be 

a 

expressed in the form ad = bx, by the law of the means and extremes. 
Since the products ad and bx are equal the whole expression, ad = bx, 
is called an equation. The quantities on the left side of the equality 
sign constitute the first member; those on the right side constitute 
the second member. Any quantity or group of quantities which is sepa- 
rated from others by the + or —sign is called a term of the equation. 
The equation has the following properties: 

(a) If letters occur in an equation, they must be given such 
numerical values that when substituted in the equation both members 
icill be numerically equal. In the equation 2b = 4A, let b = 4 and 
k = 2. Substituting these values, the equation becomes 2X4 = 
4X2, or 8 = 8. 

(b) The same quantity may be added to or subtracted from each 
member icithout destroying the equality. In the equation, 26 = 4A, let 
8 be added to each member. Thus, 2b + 8 = 4Jc + 8. Then sub- 
stituting the values given in (a) in the above equation, it becomes 

(2 x 4) + 8 = (4 X 2) + 8 

8+8=8+8 

10 = 10 

It is seen that the quantities within the parentheses, which represent 

the values of the members of the original equation, are equal and the 



90 PRACTICAL MATHEMATICS 

final result shows that the equality has not been destroyed by adding 
the same quantity to both members. 

Similarly it can be shown that the same quantity may be 
subtracted from both members without destroying the equality. 

(c) Each member of the equation may be multiplied or divided 
by the same quantity without destroying the equality. For example, 
divide by 3 each side of the equation 3a; + 6c = 9m + 15. 

3.t + 6c _ 9?n + 15 

3 ~ 3 

Let x = 25, c= 5, and m = 10. Substituting these values in the 

.. ... (3 X 25) +(6X5) (9 X 10) + 15 

equation, it becomes — — — = 

o o 

75 + 30 ^ 90 + 15 

3 3 

105^105 

3 3 

It is seen that the equality has not been destroyed by the 
division. Similarly it may be shown that the same root of each 
member may be extracted, or that each member may be raised to 
the same power without destroying the equality. 

(d) When a term has no sign before it, the plus sign is understood. 

(e) The order of the terms or of the letters ivithin the terms is 
unimportant; the proper sign, however, must be given to each term. 
For example 6a&c+4c?/ = is the same as 6cab+fyc = 0. Again, 
3a — 66c — xy = 40 is equivalent to — xy-\-3a — 66c = 40. 

71. Transposition, (a) Suppose it is required to solve lor the 
unknown value of x in the equation 4a; +2 = 6+ 3a;. In this equation 
the unknown quantity is found in both members. In any equation 
containing the unknown quantity in both members it is common to 
arrange the unknown values on the left side of the equation, and the 
known values on the right. After combining the terms divide both 
sides of the equation by the coefficient of the unknown quantity. The 
above operation may be accomplished by subtracting 3a; and 2 from 
both members of the equation which results in the following : 
4a; + 2 - 3a; - 2 = 6 + Zx - Zx - 2 

combining terms x + = 4 + 

x = 4, Ans. 






PRACTICAL MATHEMATICS 91 

Again, solve for x in the equation 

Sx + 4.r + 6 = 2x + S 

Subtracting 2x and 6, 

from both members, S.r + 4.r + 6 — 2x — 6 = 2x + 8 — 2x — 6 

combining terms 10.r = 2 

or .r = .2 Ans. 

(b) The above form of solution may be simplified by the use 
of a method known as transposition. All terms containing the 
unknown quantity, which are found on the right side, are transposed to 
the left, at the same time changing their signs; likewise all known terms 
found on the left side arc transposed to the right, at the same time changing 
signs. If, when the terms have been combined, x is found to be — , 
change its sign to + and at the same time change the sign of its 
numerical equivalent. (See Examples 6 and 8 following.) 
Examples. 1. Solve the equation, 4.r + 2 = 6 + 3.r. 
Solution: Bring over the Sx to the left-hand side of the equation 
and change its sign from + to — ; also transpose the 2 to the right- 
hand side and change its sign to — ; thus, 

Transposing 4.t — 3x = 6 — 2 

Combining x = 4 Ans. 

2. Solve the equation: 8x + 4# + 6 = 2x + 8 
Transposing 8.t + 4x — 2x = 8 — 6 

10z = 2 

2 

x = — or .2 Ans. 

3. Solve for m in the equation, 10m + 12 — 2m = 24 — 2m 
Transposing 10m — 2m + 2m = 24 — 12 

Combining 10m = 12 

vi = — = 1.2 Ans. 

4. Solve the equation : 9.t = 6.t+20 
Transposing 9x — 6.r = 20 
Combining Zx = 20 

a: = — or6§ Ans. 



92 PRACTICAL MATHEMATICS 

5. Solve the equation : 4a: — 3 = 61 
Transposing 4a: = 61 +3 
Combining 4a: = 64 

«-?«16 An, 

4 

6. Solve the equation : 8a; — 22 = 12a; — 18 
Transposing 8a:- 12a: = 22 -18 
Combining — 4a; = 4 

Dividing each side by 4 and changing signs, 

x= —1 Ans. 

7. Solve the equation : 32a; + 24 = 30 - 50a; 
Transposing 32a; + 50a; = 30-24 
Combining 82a; = 6 

6 3 A 
^82° r 41 AnS ' 

8 . Solve the equation :12 — 13a; = 5 — 10a; 
Transposing — 13x + 10a; = 5 — 12 
Combining — 3a; = — 7 
Changing signs 3a; = 7 

7 
3 






Ans. 



PROBLEMS FOR PRACTICE 

Solve for x in the following equations : 

1. 5x+9 = 14-2a; 

2. 6a;-28 = 15x-16 Ans. -l£ 

3. 20 -14a; = 26 -18a; 

4. 12x+25-35 = 14a;+22a;-22 

5. 81 + 10-20a: = 60+40a; 

72. Equations Containing Fractions. The equations which 
have so far been given illustrate more simple algebraic forms. In 
practice, equations may often contain fractions and it will be well 
to study the method of handling such equations. 

In the treatment of fractions, Part II, it was shown that a group 
of fractions could be added or subtracted only by reducing them all 
to a common denominator. This same process may be applied to 



PRACTICAL MATHEMATICS 93 

an equation containing fractions. For example, let us take the equa- 
tion 

^ + £ = 13 

7 o 

By inspection the L. C. D. is seen to be 42. Reducing each term of 
the equation to the least common denominator, we have 

(6X4.r) + (0X7) (7x.r) = 42X13 

6X7 " 7X6 42 

or 

24.r+42 7.f = 546 

42 42 42 

r ,. . (24*+ 42 + 7.r) 546 
Combining - - = — 

It has just been learned in article 70 that each member of an 
equation may be multiplied by the same quantity without destroy- 
ing the equality. Multiplying the above equation by 42, we have 

(24.r+42 + 7a) _ ^6<546 

W~ 

Cancelling like terms in numerator and denominator of each fraction, 
we have 

24.r+42+7.r = 546 
31a; =504 

x = 16.2+ 

An inspection of the last fractional equation shows that mul- 
tiplying by 42 was unnecessary as the same result would have been 
obtained by simply dropping the common denominator from each 
term. This leads to the following rules: 

Rules, (a) To solve an equation containing fractions, reduce all 
terms of the equation to a common denominator in the usual manner 
and then drop the denominators (This process is termed "clearing of 
fractions".) The resulting equation is then solved for the unknown 
quantity by the usual process. 

(b) If fractions occur in the numerator or denominator of any 
term, this term must be reduced to a simple fraction before finding the 
L. C. D. 



94 PRACTICAL MATHEMATICS 



6a: 2 

Examples. 1. Solve the equation — — (7+a:)=- 

8 3 



lS.r 24 (7+.r) 16 





~" 24 24 24 


>pi 


ng denominator 18a*— 168 — 24a* = 16 




-6a: = 184 




6a; =-184 




'a: =-30.6 + Ans 


2. 


Q i +1 .. 8a-+3 (2-4z) Q 

bolve the equation = 3 

15 5 



Clearing of fractions 8a:+3 — 6+12a: = 45 
Combining 20a: = 48 

a: = 2.4 Ans. 









EXAMPLES 


FOR 


PRACTICE 


Sol 


ve\ 


! or x in the following equations 








1. 


x 2_ 

20 4~ 


4 
= 5 










2. 


(8a-+3) 
12 


+ 1=8 




Ans. 


10| 




3. 


42-a: 


.(2,+14)_ 20 




Ans. 


—82.16 + 



25 10 

73. General Application of Equations to Engineering Problems. 

In the preceding problems the relation of the quantities has been 
expressed directly in equation form and solutions given for the value 
of the unknown quantity. However, the equation finds its greatest 
field of usefulness in problems in applied science and engineering 
where no equation is given directly but must be formed from condi- 
tions stated in the problem itself. 

The solution of such a problem consists of two distinct parts : 
(1) The statement of the problem, and (2) the solution of the 
equation. 

The statement of the problem is the process of expressing 
the conditions of the problem in the form of an equation. The 
statement of the problem is often more difficult to beginners 
than the solution of the equation. No rule can be given for the 



PRACTICAL MATHEMATICS 95 

statement of every particular problem. Much must depend on the 
skill of the student, and practice will give him readiness in this 
process. The following is the general plan of finding the equa- 
tion: 

(a) Study the problem, to ascertain what quantities in it are 
known and what are unknown, and to understand it fully, so as to be 
able to prove the correctness or incorrectness of any proposed answer. 

(b) Represent the unknown quantity by x, and express in equa- 
tion form the relations which hold between the known and unknown 
quantities; an equation will thus be obtained, which can be solved by the 
methods already given, and from which the value of the unknown quan- 
tity may be found. 

The problems which follow the illustrative examples will test 
the ability of the student to state the problem in equation form, and 
solve for the unknown quantity. 

Examples. 1. A, B, and C receive $1285 among them; A's 
share is 825 more than |ths of B's, and C's is Aths of B's. Find the 
share of each. 

In this problem we must first determine what unknown quan- 
tity is to equal x. Since both A's and C's shares are compared with 
B's, (or depend upon B's), we may let x represent B's share. Now, 
A's share is 825 more than fths of B's, and as B's share is x, A's 

must be (I Xx) 4- 25, or — — f-25. As C's share is V^ths of B's, it will 

6 

be n Xa; or — . 
15 

We now have a statement of all three shares, and since together 

the men have $1285, the sum of their shares must equal SI 285. We, 

therefore, place the sum of the three shares as the first member of 

the equation, and the actual value of these three shares in dollars 

for the second member of the equation, as follows: 

Solution. Let x represent B's share 

•5 E 
then — — [-25 represents A's share 

Ar 

and jr represents C's share 



Forming the equation, we have: 



* + (lf +25 ) + 



15 



9G PRACTICAL MATHEMATICS 

Clearing equation of fractions: 

30x+25x +750 +8z = 38,550 
Transposing: 

30.c+25.r+Sz = 38,550-750 
Combining terms: 

63^ = 37,800 

x = $600, B's share 

Since we have B's share, to find the other two shares, we substitute $600 
for x, as follows: 

A's share = ^+25 = (5 * 6 ° 0) +25 = $525 
6 b 

n> u 4x 4X600 ___ 

C's share = — = — — — =$160 
15 15 

fA's share $525 

Answers ] B 's share $600 

iC's share $160 

2. Divide 19 into two parts such that 7 times the less shall 
exceed 6 times the greater by 3. 

Solution. Let x = the lesser part 

(19 — x) = the greater part 

then 7x = 7 times the lesser part and 6 (19 — x) = 6 times the 

greater part, but 7 times the lesser part must exceed 6 times the 

greater by 3. 

Therefore, 7z-6 (19 -a;) =3 

Expanding, 7x— 114+6;r = 3 

Transposing, 13x= 114 + 3 

13z= 117 

117 
Dividing by 13, x = -—- = 9 Ans. 

lo 

Therefore, x = 9, the lesser part; 19 — 9 = 10, the greater part. It 
can readily be seen that the conditions have been met, as 7 times 
the lesser, 63, exceeds 6 times the greater, 60, by 3. 

3. In an ingot of brass weighing 105 lb., if the copper used 
weighed 20 lb. less and the tin 20 lb. more, the weight of the copper 
would still be 5 lb. more than the weight of the tin. What is the 
weight of each? 

Solution. Let x represent weight of copper and (105— a;) repre- 
sent weight of tin. 



PRACTICAL MATHEMATICS 97 

Then by the conditions of the problem, the weight of the copper 
reduced by 20 lb., that is (x— 20), is 5 lb. more than the weight of 
tin increased by 20 lb., that is, (105 — .r) + 20. The equation is, 
therefore, 

(.i— 20) -5 = (105 -.r) +20 
Transposing x + x = 105 + 20 + 20 + 5 
2x = 150 
x = 75 lb. weight of copper "1 . 
105 - 75 = 30 lb. weight of tin J nS * 

PROBLEMS FOR PRACTICE 

1. A bar of solder weighs 51 lb. It is composed of tin and 
lead. There are 9 lb. more tin in the bar than lead. How much 
lead is there? How much tin? 

(Suggestion: Let x = the amount of lead. Then 51 — x = the 
amount of tin.) 

Ans. 30 lb. tin. 21 lb. lead. 

2. Divide a freight train of 76 cars into two trains so that 
twice the longer train shall be less by 16 than 4 times the shorter 
train. 

(Suggestion: x = number of cars in longer train; 76 — x = 
number of cars in shorter train.) 

Ans. 48, 28. 

3. Two dynamos furnish 84 horsepower. One dynamo 

furnishes f as much power as the other. What does each furnish? 

3.r 
(Suggestion: x = horsepower of one machine, -— = horse- 

4 

power of other machine.) 

Ans. 48 h.p., 36 h.p. 

4. In a ton of prepared mortar, the sand weighs 200 pounds 
more than the gypsum and the plaster of Paris one-fourth as much 
as the gypsum. What is the weight of each? 

5. A tank can be filled by two pipes, A and B, in 12 and 15 
minutes respectively. A service pipe C will empty the tank in 10 
minute-. How long will it take A and B to fill the tank when C is 
emptying it! 

Ans. 20 minutes. 



PRACTICAL MATHEMATICS 

PART IV 



PRACTICAL GEOMETRY 

Geometry, the science of space, treats of lines, triangles, etc. 

A figure of one dimension, length is called a line; a figure of two 
dimensions, length and breadth, is called a swrfaee or area; for example, 
a triangle, a circle; a figure of three dimensions, length, breadth, and 
thickness, is called a solid; for example, a cube, a sphere, a cone. 
Such figures are called geometrical figures. 

LINES 

74. A line has length only; it is made by the motion of a point. 

A straight line is one that has the same direction throughout. 
It is the shortest distance between two points. 

A curved line is one that is constantly changing in direction; it 
l- -ometimes called a curve. 

A broken line is one made up of several straight lines. 

A plane surface (or simply a plane) is a surface such that any 
straight line drawn in it lies wholly within the plane. 



Straight Line. Curved Line. Broken Line. Parallel Lines. 

Parallel lines are lines which lie in the same plane and are 
equally distant from each other at all points. 

Lines are lettered to distinguish them; thus if one end is marked 
a and the other b, it is called "the line a&" or "the line 6a." 

ANGLES 

75. An angle is the measure of the difference in direction of 
two lines. The lines are called sides, and the point of meeting, the 
vertex. The size of an angle is independent of the length of its sides. 

If one straight line meets another (not at an extremity), Fig. 9, 
so the two angles formed are equal, the lines are said to be perpen- 
dicular to each other and the angles formed are called right angles. 



100 PRACTICAL MATHEMATICS 

An acute angle is less than a right angle, Fig. 10. 

An obtuse angle is greater than a right angle, Fig. 11. 

Angles are distinguished by placing letters at the end of the 
lines forming the sides and at the vertex. For example, in Fig. 10 
the angle would be read as "the angle A B C" or "the angle C B A." 
In writing, the sign Z is often used in place of the word "angle." 



B n B 



C 




Fig. 9. Right Angle. Fig. 10. Acute Angle. Fig. 11. Obtuse Angle. 

As two right angles are formed when one line meets another 
perpendicularly, Fig. 9, it follows that: 

(a) The sum of all the angles about a point on one side of a straight 
line is equal to two right angles. 

(b) The sum of all the angles about any point is equal to four right 
angles. 

When two lines intersect they form four angles, Fig. 12. The 
angles A O.C and A O D are called adjacent angles because they 

have the same vertex O and the line A O 
is common to both as it forms a side of 
each angle. It has been proved that Z 
A O C + Z AO D = two right angles and 
it may also be proved that the angles AO C 
intersecting Lines, and DOB, called opposite angles, ; are 
equal to each other. Therefore, when two 
straight lines intersect, the opposite angles formed are equal to each other. 
76. Measurement of Angles. It has been shown in Fig. 12, 
that the sum of the angles about the point of intersection of two lines 
is always four right angles. Evidently, other lines might be drawn 
through the same intersection, making the angles smaller and more 
in number, without changing the value of their sum. In measuring 
angles, therefore, it has been agreed to divide these four right angles 
about a point into 360 equal parts called degrees, indicated by the 
sign, °. In this case each of the four right angles would contain 




PRACTICAL MATHEMATICS 



101 



90° 




90° while one-half, two-thirds, and 
one-third of a right angle would rq>_ 
resent angles of respectively 45° J 
60°, and 30°, as shown in Fig. 13. 
In Drawing Work, it will be found 
that the 45° triangle and the 30° 
— 60° triangle are useful articles 
of equipment, but when it comes 
to accurate values of angles, some 
device like a protractor, Fig. 14, 
showing degrees and fractions of de- 
grees, is necessary. Each degree is 
divided into 60 equal parts called minutes, ', and each minute into 
60 equal parts called seconds, ". Thus to represent an angle of 23 
degrees, 47 minutes, and 9 seconds, write 23° 47' 9". 

The relations of these units may be summarized as follows: 

60 seconds (") = 1 minute (') 

60 minutes = 1 degree (°) 

360 degrees = 1 circle 

Two angles are complementary 
when their sum is equal to one right 
angle or 90°; they are supplement wry 
when their sum is equal to two right 
angles or 180°. Thus, an angle of 




Fig. 14. Protractor. 



33° is the complement of one of 57° because 33 -f 
of 148° and 32° are supplementary because 148 

PROBLEMS FOR PRACTICE 

1 . How many seconds in 180 degrees ? 



57 = 90. Angles 
f 32 = 180. 



Ans. 648,000 
Ans. 2} 



Ans. 78° 12' 57" 
Ans. 48° 55' 33" 



2. How many right angles in 202° 30'? 

3. "What is the complement of 37°? 

4. What is the complement of 11° 47' 3"? 

5. What is the supplement of 131° 4' 27"? 

6. Does an acute angle contain more or less than 90°? An 
obtuse angle? 

7. In Fig. 12, the angle A C is, say, 52°; find the value of the 
three other angles about the point 0. 

8. How many seconds in 30'? 



102 



PRACTICAL MATHEMATICS 
POLYGONS 



77. A polygon is a plane figure bounded by straight lines. 
The boundary lines are called the sides and the sum of the sides is 
called the 'perimeter. 




Fig. 15. Pentagon. 




Fig. 16. Hexagon. 



Fig. 17. Octagon. 



Polygons are classified according to the number of sides. 
A triangle is a polygon of three sides. 
A quadrilateral is a polygon of four sides. 
A pentagon is a polygon of five sides, Fig. 15. 
A hexagon is a polygon of six sides, Fig. 16. 
An octagon is a polygon of eight sides, Fig. 17. 
An equilateral polygon is one all of whose sides are equal. 
An equiangular polygon is one all of whose angles are equal. 
A regular polygon is one all of whose angles and all of whose 
sides are equal. 

TRIANGLES 

78. A triangle is a polygon enclosed by three straight lines called 
sides. The angles of a triangle are the angles formed by the sides. 

A a 




Fig. 18. Right- Angled 
Triangle. 



Fig. 19. Acute- Angled 
Triangle. 



Obtuse- Angled 
Triangle. 



A right-angled triangle, often called a right triangle, Fig. 18, 
is one that has a right angle. The longest side (the one opposite the 
right angle) is called the hypotenuse and the other sides are some- 
times called legs. 

An acute-angled triangle, Fig. 19, is one that has all of its angles 
acute. 

An obtuse-angled triangle, Fig. 20, is one that has an obtuse angle. 



PRACTICAL MATHEMATICS 



103 



An equilateral triangle, Fig. 21, is one having all of its sides equal. 

An equiangular triangle is one having all of its angles equal. 

An isosceles triangle, Fig. 22, is one two of whose sides are equal. 

A scalene triangle, Fig. 23, is one no two of whose sides are equal. 

(a) The base of a triangle is the lowest side; it is the side upon 
which the triangle is supposed to stand. Any side may, however, be 
taken as the base. In an isosceles triangle, the side which is not one 
of the equal sides is usually considered as the base. 






Fig. 21. 
Equilateral Triangle. 



Fig. 22. 
Isosceles Triangle. 



Fig. 23. 
Scalene Triangle. 



(b) The altitude of a triangle is the perpendicular drawn from the 
vertex to the base A D, Fig. 24. In some triangles, Fig. 25, it is 




necessary to produce the base so that the altitude may meet it. 
In a right triangle one leg may be considered as the base and the 
other as the altitude, Fig. 26. 

Rule: The sum of the angles of any 
triangle is equal to two right angles. This 
simple relation may be proved as follows: 
Let the base B C of the triangle in Fig. 27 
be extended or produced to E, and the 
line C D be drawn parallel to A B. This construction makes the 
ZDC E equal to the Z A B C, and the Z B A C equal to the ZACD. 
Forming an equation from these two equalities and adding the 
Z A C B to both sides, the following is obtained: 

ABC+BAC+ACB=DCE+ACD+ACB 




104 



PRACTICAL MATHEMATICS 



But it will be noticed that the left member of this equation com. 
prises the three angles of the triangle ABC and that the right mem- 
ber includes all of the angles about the point C on one side of the 
line B E, which have already been found to equal two right angles. 
Hence, if 

DCE + ACT) + ACB = two right angles 
then 

ABC + BAC + ACB = two right angles 
Therefore the sum of all the angles of any triangle is equal to two 
right angles. 

(c) Laiv of the right triangle. If in the right triangle ABC, Fig. 
28, the side A B is made 3 inches long, and the side A C is made four 
inches long, then the side B C, called the hypotenuse, is found to be 5 
inches long. 

The proof of this may be found by considering the accompanying 
figure. The square ABED constructed on the side A B contains 





Fig. 28. 



D £ 

Graphical Proof of the Law of the Right Triangle. 



9 squares ; the square L C AM constructed on the side A C contains 
16 squares, and the square C F OB constructed on the hypotenuse 
B C contains 25 squares. Now, 

9 + 16 = 25 
324. 42= 52 



AB +AC =BC 
Rule: The square of the hypotenuse is equal to the sum of the 
squares of the other two sides. This statement is only true for right 
triangles, and hence must not be applied to triangles in general. 
By means of this relation it is possible to £nd the length of one side 
of a right triangle if the other two are known. 



PRACTICAL MATHEMATICS 



105 



Examples. 1. A right triangle has sides 6 inches and 10 inches 
long respectively; what is the length of the hypotenuse? 
Solution. Let x represent the hypotenuse. 
x a = 6 2 + 10 2 

= 36 + 100 = 130 
• x = Vl3G = 11.66 + inches. Ans. 
2. The hypotenuse of a right triangle is 24 feet long and one 
of the short sides is 9 feet. How long is the other side? 
Solution. Let x represent the other side. 
24 2 = 9 2 + x 2 
x* = 24 2 - 9 2 

= 576- 81 = 495 
x = V495 = 22.248 + feet. Ans. 
79. Relation of the Angles and Sides of a Triangle. In any 
triangle there are six parts — three sides and three angles. Three 
sides, two sides and one angle, or one side and two angles will deter- 
mine the shape and size of the triangle. These relations are called 
functions of the angles, and the unknown parts of a triangle may be 
computed from the values of the known parts. 

Note. This word function is used in mathematics to denote a quantity 
which depends upon some other quantity for its value. For example, the price 
of eggs depends mainly upon the supply and demand; hence the price of eggs 
is a function of the supply and demand. The area of a circle is a function of the 
radius; the distance a train can cover is a function of speed and time. 

Let M A X, Fig. 29, 
be any acute angle. Then 
if lines be drawn from the 
points B, D, and F per- 
pendicular to A X, the 
right triangles ABC, 
ADE,sindAFGv,i\\beA 
formed. Such triangles, Fig - 29 - SimUar Triangles, 

having their angles all equal, are called similar, and their corresponding 
sides are proportional. 
Therefore : 

BC DE FG . B C DE FG 




DE _ FG 
AB AD AF and AC AE~A& 



etc. 



106 PRACTICAL MATHEMATICS 

These ratios which depend upon the size of the angle a for their 

values are functions of a. There are six such ratios for every 

triangle and for convenience they have been given names. The 

most important of these are the following: 

B C B C 

-j-ir = sine of a (written sin a) -j-~ = tangent of a (written tan a) 

"j-jt — cosine or a (written cos a) ^--^ = cotangent of a (written cot a). 

As the similar triangles AD E and A FG have also the angle a, the 
similar rotios of their sides may be expressed in the same way. Thus : 

DE . FG . AE FG 

-j-jz — sin a -j-p = sin a -j-tt = cos a -j-~ = tan a 

In other words, sin a is always the same for the same angle, and if 
A M and A N were prolonged so that an infinite number of triangles 
might be formed, sin a would still retain the same value. The same 
plan may be followed with angle b, Fig. 29, with the result that 

AC B C AC 

sin b = — g, cos b = -7-5* and tan b = p n - If these values are 
A ti An Jo L 

compared with those given for a, it will be seen that sin a = cos b 9 
cos a = sin b, tan a = cot 6, etc. As the angles a and b are comple- 
mentary, this may be considered the rule for complementary angles. 
The rules for the functions of an angle may be stated thus: 

opposite side opposite side 

sm a = -—^ — ■ tan a = — jf , . , • 

hypotenuse adjacent side 

adjacent side adjacent side 

cos a = —. cot a = =- — rr— 

hypotenuse opposite side - 

If the sides of the right triangle ABC, Fig. 30, have values of 3, 4, 

and 5, the values of the functions of a will be as follows : 

sin a = - = .6 tan a = - = .75 
5 4 

4 4 

cos a = - = .8 cot a = - = 1.33 + 
o 

The table, page 170, will show that 

the angle whose sine is .60 is about 37°. 

The values of cos, tan, and cot also Flg * 30 " o?sidef Give?. Lengths 

agree approximately with the table values for 37°. If 45° and 

30° — 60° triangles be taken, their functions are as easily obtained. 




PRACTICAL MATHEMATICS 107 

5 
Examples. 1. Given sin a = — , to find the value of the other 

lo 

side and the remaining functions. 

Solution. 

The opposite side = 5, the hypotenuse = 13. Hence the 

other side is equal to ^ 13 " - 5 : = 12. The three remaining functions 

12 5 12 .12 

are cos a = — -. tan a = — . cot a = —■. (Table gives cos-— = 23° — .) 

13 12 5 13 

2. Find the value of the sine, cosine, and tangent of 45°. 

Solution. A 45° triangle has its 
opposite and adjacent sides equal, hence 
if the hypotenuse is assumed to be 10, 
the sides may be calculated by the law 
of the right triangle, giving a value 7.07. 10 2 = X 2 + X 2 = 2x 2 

Hence sin 45° = ~= .707 & = 50 

7 07 * = 7 ' 07 + 

cos45° = ~= .707 

. <-o 7.07 , 
tan 4o = =-t>~ = 1 
/ .0/ 

A careful trial of all possible angles between 0° and 90° would 
show that the functions may have the following values: 

sin 0° = 0, and increasing angles have increasing values up to 
sin 90° = 1. (Verify sin, cos, and tan values from table, page 170.) 

cos 0° = 1, and increasing angles have decreasing values down 
to cos 90° = 0. 

tan 0° = 0, and increasing angles have increasing values up 
to tan 90° =00.* 

80. Areas of Triangles. A triangle is a surface and as such has 
two dimensions, base and altitude. Under denominate numbers the 
fact was brought out that surfaces or areas were measured by square 
measure, and that the area of the surface was proportional to the 
product of its two dimensions. If the surface is a triangle it has been 
found that the area is equal to one-half of the product of the base by 
the altitude. For example, in Fig. 24 and Fig. 25, the area of each 

triangle is - (B C X A D). Other methods might be given, but by 

the above method, most cases which are met may be easily solved, 

*Called the infinity sign, or "infinity," meaning infinitely large. 



108 PRACTICAL MATHEMATICS 

Examples. 1. Find the area of the triangle ABC shown in 
Fig. 26 if the base B C is 10 inches and the altitude A B is 10 inches. 

Solution. Area = s- (A B X B C) = ^ (10 X 10) =50 square inches. 

2. Find the area of a right triangle whose base is 50 feet and 
the hypotenuse 200 feet. 

Solution. To obtain the altitude, make use of the law of the right 
triangle, i.e., take the square root of the difference in the squares of the hypot- 
enuse and base of the triangle. 200- 50*= 40000 - 2500 = 37500. 

V37500 = 193.64+ ft. altitude of the triangle. 

Area of the triangle = ^ (50 X 193.64) = 4841 sq. ft. 

3. A ladder 40 feet long when placed on the ground 24 feet 
from the bottom of a wall reaches to the top. Determine the height 
of the wall. 

Solution. As the ladder is 40 feet long and the base is 24 feet from the 
wall, there is a right-angle triangle whose hypotenuse is 40 feet, one short side 
24 feet, and the other short side to be determined. As in the previous example 
make use of the law of the right triangle and this will give 40 2 — 24 2 = 1024. 
Taking the square root of 1024 will give 32 feet as the height of the wall. 



PROBLEMS FOR PRACTICE 

1. A triangular piece of sheet metal weighs 18 pounds. If the 
height of the triangle is 8 feet and the base 3 feet, what is the weight 
of the metal per square foot? 

2. In surveying a triangular piece of land it was found that 
two of the angles measured 44° 56' V and 31° 11' 8". Compute the 
value of the third angle. Ans. 103° 52' 48" 

3. An iron brace used in supporting a shelf is fastened to 
the wall 18 inches below the shelf and to the shelf 12 inches from the 
wall. Find the length of the brace. 

4. The area of a triangle is 24 square inches. If the altitude 
is 6 inches, find the length of the base. 

5. An iron chimney is supported by a guy wire which makes 
an angle of 63° 24' with the ground. Determine the angle between 
the chimney and guy wire. Ans. 26° 36'. 



PRACTICAL MATHEMATICS 



109 



6. The grade of a railroad is such that in a horizontal distance 
of 5,000 feet the track rises 300 feet. What is the tangent of the 
angle between the track and the horizontal? 

7. If one end of a ladder 50 feet long rests on the ground 10 
feet from a wall, what is the cosine of the angle formed between 
the ladder and the ground? Find the sine and tangent of the same 
angle. Answers: .2; .98; 4.9. 

8. Find the area of the right triangle of Problem 7. 

QUADRILATERALS 

81. A quadrilateral is a polygon bounded by four straight lines, 
as Fig. 31. 

c 




Fig. 31. Quadrilateral. Fig. 32. Trapezoid. Fig. 33. Parallelogram. 

A trapezoid is a quadrilateral having two sides parallel, Fig. 32. 
The parallel sides are called the bases and the perpendicular distance 
between the bases is called the altitude. 

A parallelogram is a quadrilateral whose opposite sides are 
parallel, Fig. 33. 



B 




Fig. 34. Rectangle. 



I'lg. 35. Square. 



Fig. 36. Rhombus. 



The three important kinds of parallelograms are as follows: 

The rectangle, Fig. 34, whose angles are right angles. 

The square, Fig. 35, all of whose sides are equal and whose angles 
are right angles. 

The rhombus, Fig. 36, whose sides are equal but whose angles 
are not right angles. 

82. Areas of Quadrilaterals. Considering first the trapezoid, 
it will be seen that in Fig. 32 the dotted line A C (called a diagonal) 
divides the figure into two triangles, ABC and A C D, whose 



110 PRACTICAL MATHEMATICS 

altitudes are the same, viz, the altitude of the trapezoid. By a 
previous rule the areas of these triangles will be respectively 

- (altitude X B C) and - (altitude X A D) 

7? C I AT) 
Therefore Area of trapezoid = altitude X - 



2 

Rule: (a) The area of a trapezoid is equal to the product of the 

altitude by one-half the sum of the bases. 

The area of a quadrilateral like that snown in Fig. 31 may be 

obtained by drawing a line B E, parallel to the base A D. The 

figure will then be divided into a trapezoid and an extra triangle 

whose areas are measurable by the methods given. 

The parallelogram, Fig. 33, may also be divided by means of a 

.- . . . -ii • i A D X altitude 
diagonal into two triangles whose areas are respectively 

and B C X altitude But ^ D and B C are equal and, consequently, 

2 

the sum of these areas = — - = A Dx altitude. 

Li 

Rule: (b) The area of a parallelogram is equal to the product of 
the base and altitude. This will be true of any parallelogram and hence 
true of the rectangle, Fig. 34, the square, Fig. 35, and the rhombus, 
Fig. 36. In the case of the first two, the fact that the angles are all 
right angles makes the length of the vertical side become the altitude. 

Rule: (c) The area of a rectangle is equal to the product of two 
adjacent sides. This has already been learned in denominate num- 
bers. The fact that all land measurements are made in the rectan- 
gular system, makes the rectangle an extremely important figure. 

Areas of Other Polygons. In finding 
the areas of polygons, other than the 
triangle and quadrilateral, the system of 
dividing the figure into triangles is used. 
For example, the regular hexagon, Fig. 37, 
consists of six equal equilateral triangles, 
and consequently the area of the hexagon 

will be 6 ( AB * 0X ) = 3 (4 B x o L). Fig ' 37 - ^ 6xas ° n - 




PRACTICAL MATHEMATICS 111 

"Rule: (d) The area of a regular hexagon is equal io three times the 
product of one side (which is the same as § the perimeter) by the per- 
pendicular distance from the side to the center of the hexagon. 

PROBLEMS FOR PRACTICE 

1. What is the area of a cross-section of a regular hexagonal 
drain tile whose sides are 2 inches in length and whose perpendicular 
distance from the center to a side is 1.7 inches? 

2. How much would it cost to lay a concrete sidewalk 8 feet 
wide and 525 feet long at 20 cents per square foot? 

3. How many square yards of plastering are needed for walls 
and ceiling of a room 10 X 12 feet and 9 feet high; the room contains 
a door 3J X 7 feet and two windows 3i X 6 feet? Ans. 50 — sq. yd. 

4. An irregular shaped room has two parallel sides, one 14 
ft. 8 in. long, the other 20 ft. 3 in. long. One end of the room is 
perpendicular to the two sides and is 12 ft. 5 in. long. What will 
it cost to paint the floor at 25 cents per square yard? Ans. $6.02+ 

5. A trough two feet deep with equally sloping sides is con- 
structed. One end is 5 feet across the top and 4 feet across the bottom. 
How many square feet of tin will be required for the end? 

CIRCLES 

83. A circle is a plane figure bounded by a curved line called 
the circumference, every point of which is equally distant from a point 
within called the center, Fig. 38. 

A diameter of a circle is a straight line drawn 
through the center, terminating at both ends in 
the circumference. 

A radius of a circle is a straight line joining 
the center with the circumference. All radii of 
the same circle are equal and their length is 
always one-half that of the diameter. FI s- 38. Circle. 

An arc is any part of the circumference of a circle. 

An arc equal to one-half the circumference is called a semi- 
circumference. 

A chord is a straight line joining the extremities of an arc, Fig. 
39. AYhen a number of chords form the sides of a polygon, the 
polygon is said to be inscribed in the circle. 




112 



PRACTICAL MATHEMATICS 




Fig. 39. Chord and 
Tangent. 



A segment of a circle, Fig. 40, is the area included between an 
arc and a chord. 

A sector is the area included between an arc and two radii drawn 
to the extremities of the arc, Fig. 40. 

A tangent is a straight line which touches 
the circumference at only one point, called the 
point of tangency or point of contact, Fig. 39. 
It may be proved that a tangent is perpendic- 
ular to a radius drawn to the point of tan- 
gency. 

Concentric circles are circles having the same 
center, Fig. 41. 

Relation of the circumference to the diameter. If the lengths of 
the circumference and the diameter of a circle are carefully measured, 

the ratio, - , will be found to have a 

diameter 

fixed value, whatever circle may be selected. This 

constant ratio has been given the name tr (Greek 

letter pi) and has a value of 3.1416 (approx- 

22, 
imately — ) • Therefore, the circumference of any 

circle is equal to it times the diameter. In most 

22 

cases the value — may be used with safety as the error is only 

about .04%. Then, to find the circumference when the diameter is 

22 
known, multiply the diameter by -=-. Conversely, 

to find the diameter when the circumference 

22 
is known, divide the circumference by -=-. 

84. Areas of Circles. Let a regular hexagon 
be inscribed in a circle, as shown in Fig. 42. 
It has already been found that the area of the 
hexagon is equal to one-half of OL times the sum of the 6 sides, 
i. e.y times the perimeter. 

Now if the number of sides of the inscribed polygon should 
be increased, the perimeter would approach nearer to the circum- 




Fig. 40. Segment 
and Sector. 




Fig. 41. Concentric 
Circles. 






PRACTICAL MATHEMATICS 



113 



^A — ' 7v\ 

At- ~7v7 



Fig. 42. Inscribed Hexagon. 



ference and the line OL would approach 

the radius of the circumscribing circle. 

Therefore, if the number of sides should 

be increased indefinitely, the areas of the 

polygon and circle would be one and the 

same and equal to one-half the radius 

times the circumference. 

. radius . 

Area of circle = — - — X circumference. 



But the circumference of the circle = tt X diameter = t X 
twice the radius. Hence 

Area of circle = I -^— — I X (2xX radius) = t (radius) 2 . 



or 



Area of circle 



X (diam.) 2 = .7854 (diam.) 2 



Rule: The area of a circle is equal to the square of the radius 

22 

multiplied by ~; or is equal to the diameter squared times .7854. 

85. Areas of Sectors and Segments. The area of a sector is 
found by determining its fraction of the area of the corresponding 
circle. Thus, if the sector has an arc of 60°, the area of the sector 
will be 60-^360 = one-sixth of the area of the circle, which has a 
radius equal to one of the sides of the sector. Similarly a sector 
with an arc of 45° has one-eighth the area of the corresponding circle. 

There are many rules and tables for 
finding the area of a segment. The follow- 
ing formula is very simple and gives approx- A 
imate results. The derivation of this for- 
mula is very difficult, as it involves higher 
mathematics; the formula may be used, 
however, by students and engineers who 
wish a simple expression giving approx- 
imate results. 




Fig. 43. Area of a Segment. 



Area (^4) of segment = — - \ -r~- 



.608 



In this formula h 
circle, Fig. 43. 



height of segment and D = diameter of 



114 PRACTICAL MATHEMATICS 

Example. 1. Find the area of a segment when the diameter of 
the circle is 12 feet and the height of the segment is 4 feet. 



A _4X4 2 1 12 

A — 3-\T" 

Solution. In this case h =4 feet 64 

and D = 12 feet. Substituting these - ~7r VS —.608 

values in the formula, the result obtained 

is 33 sq.ft. =^-1/1392=^X1.55 

= 33 + square feet 

It has been shown that a line has but one dimension and a sur- 
face or area, two. If this be true, then any expression representing 
a length or area will always contain respectively one or two dimensions. 
For example, the expression for the length of the circumference of 
a circle is ir D, which is the product of tt, a pure number, and the 
diameter of the circle, a dimension. Again, the expression for the 
area of a circle is tt R 2 , in which the radius occurs twice giving the 
two dimensions necessary for an area; similarly, the expression for 
the area of a triangle (J base X altitude) shows two dimensions. 
This fact is of use in proving up any expression. For instance, the 
expression given above for the area of a segment, could be judged as 
an area formula at once by noting that the dimension h occurs twice 
outside of the radical; and the dimension D in the numerator and h 
in the denominator of the fraction under the radical cancel each other. 

PROBLEMS FOR PRACTICE 

1. The diameter of a pulley is 18 inches. What is the circum- 
ference? 

2. The bottom of a tank is circular and is 3 feet in diameter. 
What is the area of the bottom? 

3. What is the area of one side of a flat circular ring whose 
outside diam. is 8 inches and inside diam. 4 inches? Ans. 38 — sq. in. 

4. A carriage wheel 4 feet in diameter makes how many revolu- 
tions in traveling 1,000 feet? 

5. The circumference of a chimney of circular cross section 
is 47.12 feet. What is its diameter? Ans. 14.9-J- ft. 



PRACTICAL MATHEMATICS 



115 



6. In patching the circular bottom of a steel tank 6 feet in 
diameter, a sector is cut out, the length of whose arc is 3| feet. Find 
the area of the patch required, disregarding the overlapping of the 
edge-. Ans. 4.71 + sq. ft. 

7. The bottom of a cylindrical cistern 9 feet in diameter is to 
be tinned. If sheets of tin 3 feet wide are available, find the areas of 
each of the 3 pieces used. The two end pieces have the shape of 
segments. 

8. Find the number of square feet of glass required for a 
circular arched window whose height is 10 inches and the radius 
of whose arc is 24 inches. Ans. 1.89 + sq. ft. 

SOLIDS 

86. A solid has three dimensions — length, breadth, and thick- 
ness. The most common forms of solids are prisms, cylinders, 
pyramids, cones, and spheres. 

PRISMS 

87. A prism is a solid having two opposite faces, called bases, 
which are equal and parallel, and other faces, called lateral faces, 
which are parallelograms. The altitude of a prism is the perpendicu- 
lar distance between the bases. 

Prisms are called triangular, rectangular, hexagonal, etc., accord- 
ing to the shape of the bases. Further classifications are as follows: 



<_!> 





Fig. 44. Right Prism. Fig. 45. Parallelopiped. 



Fig. 46. Rectangular 
Parallelopiped. 



A right prism is one whose lateral faces are perpendicular to 
the bases, Fig. 44. 

A regular prism is a right prism having regular polygons for 
bases. 

A parallelopiped is a prism whose bases are parallelograms, 
Fig. 45. If all the edges are perpendicular to the bases, it is called 
a right parallelopiped. 



116 



PRACTICAL MATHEMATICS 



A rectangular parallelepiped is a right parallelopiped whose 
bases and lateral faces are rectangles, Fig. 46. When these faces 
are all squares the prism is called a cube. 

88. Areas of Prisms. The area of the bases of any given prism 
is to be found by the method already given for that particular shape of 
base. The lateral area, as each face is a parallelogram, will be the 
sum of the products of each base line by the altitude. This is equiva- 
lent to the product of the perimeter of the base by the altitude. The 
total area will evidently be the sum of the areas of the two bases and 
the lateral faces. 

Example. Find the total area of a box 3 feet long, 2 feet wide, 
and 6 feet deep. 

Solution. 

The area of one base is 2 X 3 or 6 sq. ft., and of two, 12 sq. ft. 

The perimeter of the base = 2 + 2 + 3 + 3 = 10 ft.; the 
altitude = 6 ft. 

The lateral area = 10 X 6 = 60 sq. ft. 

The total area = 60 + 12 = 72 sq. ft. Ans. 

89. Volumes of Prisms. It has already been made clear in the 
discussion of Cubic Measure in denominate numbers, that the measurer 
ment of volume involves the product of the three dimensions of the 








- r ~zrr- - ~ 






Lil 


1 i 




4. — r — 

. 4. j/'i 




—y 




1 *' • 

,*' 1 




,4T 


i I 

-U-"-' 




_i . 






^^^ 


^*" 









Fig. 47. Volume of a Rectangular Solid. 

figure and is determined by the number of times the unit of cubic 
measure, such as the cubic inch, cubic centimeter, or cubic foot, is 
contained in the volume under consideration. If A, Fig. 47, rep- 
resents a cubic foot, it will be noticed that A is contained four times 



PRACTICAL MATHEMATICS 



117 



in the bottom layer of B. As there are three layers, A is contained 
a total of twelve times in B, i.e., the volume of B is 12 cubic feet. 
But the area of the base of B is 4 sq. ft. and the altitude is 3 ft. The 
product of these two gives 12 cu. ft., the same as obtained by the use 
of the unit A. Rule: The volume of any prism is equal to t lie product 
of the area of the base by the altitude. 

PROBLEMS FOR PRACTICE 

1. What is the weight of a cast-iron block 6 inches long, 5 inches 
wide, and 3 inches high? The block is a rectangular parallelopiped 
and weighs .26 lbs. per cubic inch. 

2. How many square feet of sheet copper will be required to 
make an open rectangular tank 7 feet long, 3 feet wide, and 1% feet 
deep, allowing 10% extra for waste? 

3. How many cubic feet of concrete will be required to construct 
a column 15 feet high having a hexagonal cross-section. The cross- 
section is 1J feet long on each side of the hexagon and the perpen- 
dicular distance from center to each side is 1 .3 feet. Ans. 87.75 cu. ft. 

4. What will it cost to paint the above column, exclusive of the 
base and top, with waterproofing at 35 cents per square yard? 

5. Find the volume of a triangular prism whose altitude is 20 
feet, and each side of the base is 4 feet. Ans. 138.4 cu. ft. 

6. Find the total area of the prism given in Problem 5. 

CYLINDERS 

90. A cylinder is a solid having as bases two 
equal parallel curved surfaces and as its lateral face the 
continuous surface generated by a straight line connect- 
ing the bases and moving along their circumferences. 
The bases are usually circles and such a cylinder is 
called a circular cylinder. 

A right cylinder, Fig. 48, is one whose side is 
perpendicular to the bases. 

91. Areas of Cylinders. As a cylinder may be considered a 
prism with an infinite number of sides, Us lateral area is equal to the 
product of the circumference of the base by the altitude. 

The total area is evidently equal to the sum of the areas of the 
bases and the lateral area. 



Fig. 48. 
Right Cylinder. 



118 PRACTICAL MATHEMATICS 

92. Volumes of Cylinders. The volume of a cylinder is equal 
to the product of the area of one base by the altitude. 

Examples. 1. How much tin will be required to make a 
cylindrical can, 3 inches in diameter and 4 inches high, allowing 
nothing for seams and waste? 

Solution. 

The can is a right cylinder the circumference of whose base is 

22 66 

-=- X 3 = — = 9.43 inches, nearly. Now, the radius of the 

22 
base is 3 -*- 2 = lj inches, and the area of the base — X (l£) 2 

= 7.08 sq. in., nearly. 

The lateral area is 9.43 X 4 = 37.72 sq. in. 

The total area equals 37.72 + (2 X 7.08) = 37.72 + 14.16 = 51.88 

square inches. 

2. A piece of iron weighing 30 pounds has 4 holes 6 inches 
deep and 1 inch in diameter bored into it. What is the weight of 
the piece after the boring has been completed, iron weighing .26 lbs. 

per cubic inch? 

Solution. 

1 . 22 

Each hole has a radius of — inch and its volume is then — 

X (i) 2 X 6 = 4.7 cu. in. 

There are 4 holes making 4 X 4.7 = 18.8 cu. in. 

The weight of the material cut out is 18.8 X .26 = 4.9 lbs. 

Then the final weight is 30 - 4.9 lbs. = 25.1 lbs. 

PROBLEMS FOR PRACTICE 

1. What is the capacity of a cylindrical tank 9 feet long and 
5 feet in diameter, inside measurements? 

2. If a gallon contains 231 cubic inches, what must be the 
diameter of a cylindrical tank 10 feet high which will hold 1,000 

gallons? Ans - 4- 1 + ft * 

3. Find the number of square feet of material necessary for 
a straight piece of copper pipe 10 feet long and 12 inches in diameter. 

PYRAMIDS 

93. A pyramid is a solid whose base is a polygon and whose 
sides are triangles. 



PRACTICAL MATHEMATICS 



119 



The vertices of the triangles meet to form the vertex of the 
pyramid. 

The altitude of the pyramid is the perpendicular distance from 
the vertex to the base. 

Pyramids are named according to the kind of polygon forming 
the base, viz, triangular, quadrangular, Fig. 49, pentagonal, Fig. 50, 
hexagonal, Fig. 51. 






Fig. 49. Pyramid. 



Fig. 50. Regular Pyramid. Fig. 51. Hexagonal Pyramid. 



A regular pp'amid is one whose base is a regular polygon and 
whose vertex lies in a perpendicular erected at the center of the base, 
Fig. 50. 

The slant height of a regular pyramid is a line drawn from the 
vertex perpendicular to a side of the base. (See the line F, 
Fig. 52.) In other words, it is the altitude of one of the triangles 
which form the sides. 

The lateral edges of a pyramid are the intersections of the 
triangular sides, called the faces. 

94. Areas of Pyramids, (a) The lateral area is the combined 
area of all the triangles forming the sides. 

The area of each triangle is equal to 
the product of the base by one-half the 
altitude. Therefore the lateral area of a 
pyramid is found by adding the products 
of each side of the base by one-half the 
same altitude; i.e., it is equal to the per- 
imeter of the base multiplied by one-half 
the slant height. If the slant height is 
not given it can usually be found by 
means of the law of the right triangle. 

(b) The total area of a pyramid is 
equal to the sum of the lateral area and the area of the base. 




Fig. 52. Pyramid Showing 
Altitude and Slant Height. 



120 PRACTICAL MATHEMATICS 

Example. Find the lateral area and total area of the pyramid 

shown in Fig. 52, if it has an altitude E of 12 feet and each side 

of the square base is 8 feet. 

Solution. 

E F = 4 feet. Since the angle E F is a right angle 

OF 2 =(JE + EF 

= 144 + 16 

= 160_ 

OF = Vim == 12.65 (nearly) 

rp, i , i 12.65 X 32 

1 he lateral area = 

2 

= 202.4 square feet. Ans. 

The area of the base is 8 X 3 = 64 sq. ft, making the total 
area 64 + 202.4 = 266.4 sq. ft. Ans. 

95. Volumes of Pyramids. It may be shown that any trian- 
gular prism may be divided into three equivalent triangular pyramids, 
having bases and altitudes equal to those of the prism; and therefore 
the volume of each pyramid must be one-third of the volume of the 
prism. But the volume of a prism is equal to the product of the 
area of the base by the altitude; therefore, the volume of a pyramid 
is equal to the product of the area of the base by one-third of the altitude, 

PROBLEMS FOR PRACTICE 

1. A triangular pyramid is 9 inches in height and each side of 
the base is 4 inches long. Find the volume. 

2. What is the volume of a square pyramid one side of whose 
base is 4 inches and whose height is 4 inches? 

3. Find the total area of a square pyramid whose slant height 
is 28 inches and one side of whose base is 8 inches. 

4. A regular hexagonal pyramid has an altitude of 18 inches, 
and each side of the hexagon is 6 inches long. Find total area and 
volume of pyramid. Answers: 430.84 + sq. in.; 561.16 + cu. in. 

CONES 

96. A cone is a solid bounded by a conical surface and a plane 
which cuts the conical surface. It may be considered as a pyramid 
with an infinite number of sides. 



PRACTICAL MATHEMATICS 



121 



The conical surface i-s called the lateral area, and the plane is 
called the base. The conical surface tapers to a point called the vertex. 

The altitude of a cone is the perpendicular distance from the 
vertex to the base. 

An clement of the cone is any straight line from the vertex to the 
perimeter of the base. 





Fig. 53. Circular Cone. Fig. 54. Cone of Revolution. 

A circular cone is a cone whose base is a circle, Fig. 53. 

A right circular cone, or cone of revolution, Fig. 54, is a cone whose 
axis is perpendicular to the base. It may be generated by the revolu- 
tion of a right triangle about one of the sides as an axis. 

97. Areas of Cones. The lateral area of a cone is found in the 
same way as in the case of a pyramid. Multiply the perimeter of the 
base by one-half the slant height. 

Example. The base of a cone is 12 inches in diameter and the 
slant height is 16 inches. "What is the total area? 
Solution. 

Perimeter of base = 12 X — 

7 

= 37.71 inches 
i a. 
Lateral area = 37.71 X — 

= 301.68 square inches 

To find the total area, add to this the area of the base. In the 
above example, the area of the base is 

■y X 36 = 113.14 sq. in. 

Total area = 113.14 + 301.68 - 414.82 square inches. 

98. Volumes of Cones. As a cone may be considered a pyr- 
amid with an infinite number of sides, it follows from Section 95 that, 



122 PRACTICAL MATHEMATICS 

the volume of a cone is equal to the product of the area of the base by 

one-third the altitude. 

Example. The altitude of a cone is 18 inches and the radius of 

the base is 2 inches. What is the volume? 

22 
Area of base = y X4- 12.57 

Volume = 12.57 X 6 = 75.42 cubic inches 
In case the altitude is not known, it may be found if the slant 
height and the radius of the base are known. 

PROBLEMS FOR PRACTICE 

1. Find the volume of a cone whose altitude is 12 inches and the 
diameter of whose base is 6 inches. 

2. Find the total area of a cone having an altitude of 21 feet 
and a base whose diameter is 15 feet. Ans. 702.4 sq. ft. 

SPHERES 

99. A sphere is a solid bounded by a curved surface, every point 
of which is equally distant from a point within called the center. 

The diameter is a straight line drawn through the center and 
having its extremities in the curved surface. The radius is the straight 
line from the center to a point on the surface; it is equal to one-half 
the diameter. 

A plane is tangent to a sphere when it touches the sphere in only 
one point. A plane perpendicular to a radius at its outer extremity 
is tangent to the sphere. 

100. Areas of Spheres. To find the area of the surface of a 

22 
sphere, multiply the square of the diameter by — . 

Example. Find the area of the surface of a sphere which is 9 

inches in diameter. 

22 
9 2 = 81. 81 X — = 254.6 square inches 

To find the diameter when the surface is given, divide the surface 

22 
by — and extract the square root of the quotient. 

Example. What is the diameter of a sphere whose area is 113.14 

square feet? 

99 

113.14 -7- — - = 36 nearly. >/36 = 6 feet 



7 






PRACTICAL MATHEMATICS 123 

101. Volumes of Spheres. To find the volume of a sphere, 

7T 11 

multiply the cube of the diameter by -7- (equal to — - nearly). 

b 21 

Examples. 1. A sphere is 10 inches in diameter; what is the 

volume? 

10 3 = 1000. 1000 X ij = 523.8 cubic inches 

2. Find the volume of a sphere having a diameter of 7 feet. 

7 3 = 343; 343 X ^ = 179§ cubic feet 

3. A certain sphere contains 8, 184 cubic feet. Find the diameter. 
Solution. 

To find the diameter when the volume is known, divide the volume 

by — and find the cube root of the quotient (by Logarithms).* 

11 7U 21 

8184 + ii = UU X ~ = 15624 



li 15624 = 25 ft. nearly. Ans. 

An examination of the different expressions for volume given in 
the preceding sections, in a manner similar to that given on Page 118, 
will show that each expression contains three dimensions. 

PROBLEMS FOR PRACTICE 

1. The surface of a sphere contains 314.16 square inches. 
What is the diameter? 

2. ^What will be the volume of an aluminum sphere which has 
a diameter of 15 inches? Ans. 1.02 cu. ft. 

3. How much will a sphere of cast iron weigh if it is 3 feet in 
diameter and iron weighs 450 lbs. per cu. ft.? 

4. How much will it cost to paint the sphere in Problem 3 
at 25 cents per square yard? 

5. How much less surface has a sphere which is 12 inches in 
diameter than a cube with an edge of 2 feet? Ans. 20.85 + sq. ft. 

6. A spherical balloon, when inflated, has a diameter of 50 
feet. Find the surface and the number of cubic feet of gas con- 
tained in the bag. 

♦Reserve this solution until the next section has been studied. 



124 PRACTICAL MATHEMATICS 

LOGARITHMS 

102. Methods of Calculation by Logarithms. The use of 
logarithms greatly simplifies multiplication, division, raising quan- 
tities to higher powers, and finding roots of quantities. It is not, 
however, our purpose at this time to try to explain how logarithms 
are derived, or to make clear why adding logarithms multiplies the 
numbers and subtracting logarithms divides them. Leaving this 
explanation to a later part of the course, it is sufficient at this point 
to explain carefully the method of calculation, and to give many 
examples of the manner of using the logarithmic tables on pages 
171 and 172. 

103. Characteristic and Mantissa. The logarithm of a number 
consists of two parts. One part at the left of the decimal point is 
called the characteristic and has a positive, zero, or negative value, 
depending upon the size of the number. The other part at the right 
of the decimal point is called the mantissa, and is a number depending 
wholly on the figures which form the number whose logarithm is 
desired. The characteristic of any number between 1 and 10 is 0; 
that is, such numbers as 2, 4.62, 7.845, 9.57 have a zero character- 
istic. Numbers between 10 and 100, such as 10.2, 33.42, 85.236, 
have a characteristic of 1. Numbers between .1 and 1 have a 
characteristic of —1, etc. Putting this into the form of a rule: 
Any number equal to or greater than 1 has a characteristic which is one 
less than the number of digits in the number; any number less than 1 
has a negative characteristic which is one more than the number of zeros 
between the decimal point and the first numerical figure. 

For example 

36.4 has a characteristic 1 



523 


a 


tc 


(C 


2 


532,654 


tt 


(( 


a 


5 


1.532 


(( 


a 


a 





.0072 


a 


(C 


(C 


-3 


.000038 


u 


(( 


(C 


-5 


10 


(l 


a 


n 


1 



Practice this rule by finding the characteristics of a great many 
numbers until the process is clear to you. 



PRACTICAL MATHEMATICS 



125 



The mantissa is the part of the logarithm to the right of the 
decimal point and has a certain value for every succession of figures. 
For example, the mantissa is the same for 2734, 27.34, 2,734,000 
and .002734. The value of the mantissa must always be obtained 
from the logarithm tables and may be found as follows: 

Referring to the portion of the logarithm table given below 
it will be noticed that the column of figures at the extreme left con- 
sists of two figures each from 10 to 19. 

LOGARITHMS 




Next there are ten columns of four figures each, headed by the digits 
from to 9 inclusive. These four-figured numbers are the man- 
tissas of the numbers made up of the two figures at the left and a 
third at the top of each column. For example, the mantissa of 118 
is found by taking the number 11 in the left-hand column and 
moving horizontally to the column headed 8 which is the third figure 
of the number 118; the mantissa sought is .0719. Remember that 
.0719 is also the mantissa of 1.18, 11.8, .00118, or 11,800. Again 
for 156,000, we move horizontally toward the right from 15 to the 
vertical column 6 and find the mantissa to be .1931. 

By consulting the complete tables, pages 171,172, it will be seen 
that this system is easily extended to include the mantissa for any 
succession of three numbers whatsoever. A fourth figure may be 
found by making use of the nine columns at the extreme right as 
follows: Suppose we wish to find the mantissa of 1648. The 
mantissa for 164 is found by the method . already explained and 
proves to be .2148. For the fourth place (viz, 8 in the number 1648) 
move horizontally from .2148 as far as column 8 at the extreme 
right and find 21. We must therefore add 21 to .2148 making the 
final mantissa .2169. The mantissa for 1893 is .2765 + 7 = .2772. 



120 



PRACTICAL MATHEMATICS 



This process enables us to find any succession of four figures and 
for that reason these logarithm tables are called "four-place" tables. 
Tables of six and eight places are more complicated and are only 
used for the more accurate work. It should be mentioned here that 
if the student checks the results obtained by the four-place tables 
with the actual multiplication or division of the numbers, he will 
find slight differences between these results and those obtained from 
the logarithm tables. These differences, however, are compara- 
tively small, and may be neglected as they are well within the per 
cent of error permitted in practical work. 

104. Finding the Logarithm of a Number. It only remains to 
put the characteristic and mantissa together in order to complete 
the logarithm. For example, referring to the numbers indicated 
on the portion of the logarithmic table already given, the number 
118 has a characteristic of 2 and a mantissa of .0719. The logarithm 
(or log "for short") of 118 is, therefore, 2.0719. The number 
156,000 has a characteristic of 5 (one less than the number of places) 
and its mantissa is .1931. The complete log for 156,000 is 5.1931. 
Without further analysis it may be seen that the log of 1648 is 
3.2169 and the log of 1893 is 3.2772. 

Following this process, verify the following logarithms from the 
table. 



Number 


Logarithm 


Number 


Logarithm 


3.46 


0.5391 


473,600 


5.6754 


.00856 


-3.9325 


872.6 


2.9408 


.9642 


-1.9842 


.0998 


-2.9991 


24.63 


1.3914 


6.843 


0.8353 


10,800 


4.0334 


34,970 


4.5437 



It often occurs in the use of four-place tables that the loga- 
rithms of numbers having five or more figures are desired. In this 
case increase the fourth place of the number by 1 if the fifth place 
is greater than 5, and drop the fifth and following places if the fifth 
place is less than 5. Thus the number 36.547 becomes 36.55 and 
its logarithm may be found from a four-place table. Similarly 
2.7384 becomes 2.738. 



PRACTICAL MATHEMATICS 



127 



105. Finding the Number From its Logarithm. The reverse 
process of finding the number which corresponds to a given logarithm 
is easily accomplished. Let us take the logarithm 2.7G79 and find 
the corresponding number. Consulting the portion of the logarithm 
table shown below, we find that the mantissa .7679 is the fourth 
mantissa in the column headed 6. Therefore, 6 is the third figure 
of the number sought and moving horizontally to the last column at 
the left we find 58. This means that the mantissa .7679 corresponds 



LOGARITHMS 



7404 

74S 

7559 



'709 



55 
56 
57 

59 
60 

61 
62 
63 
64 



*8©3 
7924 
7993' 
8062 



7412 
7490 
7566 



7419,7427 
7497,7505 
757475S2 



7 .i 1,; 76 H i 7t .. 'i7 



716 

389 



'931 
8000 
8069 



[23 7731 

'\)G 7803 



7868 7875 
7938 7945 
8007 8014 
S075;8082 



7435 
7513 
7589 
^64 



7738 



«*£ 



*8£ 



952 
8021 
S089 



7443 
7520 
7597 

7&a 



'45 



>4£ 



•■mi 



7959 

8028 
8096 



Y.79 



7^4 



74 59 

5 36 

■612 

'6 56 

30 



&£ 



8 



7406 
7543 
7619 
7694 
7767 



^+H 



■?$96I 7003 | 791Q 
7966 " 
8035 
8102 



7973 
8041 
8109 



980 
8048 
8116 



71 n 

7551 
7C27 
77 01 
77 74 



•846 



t9+? 
7987 
8055 
8122 



12 3 4 5 



12 2 3 4 

12 2 3 4 

12 2 3 4 

112 3 4 

112 3 4 



112 3 4 

112 4 

112 3 3 

112 3 3 

112 3 3 



6789 



^^6~ 

5 i< 6 
5 J. 6 
4 u 6 7 
4 i\ 6 



to the succession of figures 586. As the characteristic is 2 and a 
number with such a characteristic must consist of three numerical 
places, 586 is the number. Again, take the log 1.7846. The man- 
tissa .7846 is in the column headed 9, and is the mantissa for the 
succession of figures 609. As the characteristic is 1 the number 
is 60.9. 

When the mantissa of the logarithm given falls between two 
consecutive mantissas of the table, the columns at the right must be 
used. Let us find the number corresponding to the log —2.7908. 
Searching the columns no mantissa of this exact value is found; the 
next smaller mantissa is 7903, the difference being 5. Moving 
horizontally to the right from 7903, the difference 5 is found in the 
column headed 7. Therefore, 7 is the fourth place in the number 
sought and the succession of figures for the mantissa .7908 is 0177. 
As the characteristic is —2, the number sought must have one cipher 
between the decimal point and the first figure; the number is, 
therefore, .06177. 

Following this process verify the correctness of the numbers for 
the following logarithms: 



128 



PRACTICAL MATHEMATICS 



Given 
Logarithm 


Mantissa 

in Table 

Next Small- 


Difference 


Figures 
Correspond- 


Number 
Correspond- 




BB THAN 
GIVEN 




ing to 

given Log 


ing to 
given Log 




Mantissa 








4.5755 


.5752 


3 


3763 


37630 


-1.7860 


.7860 





611 


.611 


3.0000 


.0000 





100 


1000 


0.9188 


.9186 


2 


8293* 


8.293 


-3.6289 


.6284 


5 


4255 


.004255 


0.4442 


.4440 


2 


2781 


2.781 



*When two of the right-hand columns contain the same number, choose the smaller. 

106. Multiplication by Logarithms. Now, suppose we wish to 
find by logarithms the product of two numbers like 251 and 3.16. 
The mantissa of 251 is .3997; the characteristic is 2; and the com- 
plete log. 2.3997. Similarly the log. of 3.16 is found to be 0.4997. 
Adding these two logarithms we have 2.8994, which is found from 
the table to be the log. of 793.2. 

Logarithms should not always be used as they are sometimes 
not as simple as plain multiplication; for instance, 1,000 X 3.16 = 
3,160, which result can be obtained mentally, but the product of 
two such numbers as 7,853 and 522 could not be so easily calculated. 
The characteristics of these numbers are respectively 3 and 2 (in 
all cases, one less than the number of places in the whole number) and 
by consulting the table the mantissas are respectively .8951 and .7177. 
The product of 7,853 and 522 is found, therefore, by first adding 
3.8951 and 2.7177, giving 6.6128, then finding the number (410) 
whose mantissa is .6128 and finally pointing off 7 places (one more 
than the characteristic) in the result, giving 4,100,000 as the product. 

Examples. 1. Find the product of 24 and 7,635. 



Solution. The characteristics of the 
numbers are 1 and 3. The mantissas are found 
to be .3802 and .8828. Adding the logarithms 
gives 5.2630, which is found to be the loga- 
rithm of .183,200, the required product. 



Log. 
Log. 



24 
7635 



1.3802 

3.8828 



Log. 183,200 = 5.2630 



2. Find the product of 8,995 and 6,987. 



PRACTICAL MATHEMATICS 129 

Solution. As both numbers are greater 
than 1,000 and less than 10,000, their char- 
acteristics are 3. The table shows for S,99o Lo°\ S995 = 3.9540 
a mantissa = .9540, and for 6,9S7, a mantissa ho*'. 6987 = 3.8443 
= .8443. Adding the "logs" gives 7.79S3. * _ 7 ' 7QW 
The table gives 6,284 as the number whose Log.62,840,000 - 7./ 983 
mantissa is .7983, and a characteristic of 7 
makes the number 62,840,000. 

3. Find the product of 42.873 and .542. 

Solution - . As the tables give only four 
places, the last figure of the first number is ^ _ -.^-i 

neglected. If it had been 5 or more, the fourth Lo ^ 5^ = - 1 7340 

place would have been increased by 1. It »' * ' 

must be remembered that the characteristic Log. 23.23 = 1.3661 

depends only on the ichole number, while the 

mantissa disregards the decimal point entirely. 

107. Division by Logarithms. As division has already been 
defined as the reverse of the process of multiplication, division by 
means of logarithms is accomplished by subtracting the log of the 
divisor from the log of the dividend. This operation presents no 
difficulties except in case the characteristic of the subtrahend exceeds 
that of the minuend or in case a negative characteristic occurs in 
the final result. See Problem 2. 

Examples. 1. Divide 8,754 by 37.6. 

Solution. The logarithms are found 
in the same manner as in the previous exam- 1^ &- 8754 = o.y4zz 

pies, but the logs, are subtracted in this case, Log. 37.0 = 1.5752 

giving 2.3670, which is the log. of 232.8, the Log. 232.8 = 2.3670 

correct quotient. 

2. Divide 2.34 by 865. 

Solution. In this example the dividend 
is smaller than the divisor and hence the sub- 
traction results in a minus characteristic. This 
gives two ciphers to the left of the first numer- 
ical place, making the result of the division 

.002705. The minus characteristic is some- Log. 2.34 = 0.3692 

times avoided by adding 10 to the character- Log. 865 = 2.9370 

if the dividend and indicating a subtrac- Log. .002705 = — 3.4322 
tion of 10 after the mantissa; thus, 0.3692 = 
10.3602 — 10 and the subtraction now gives 
7.4322 - 10 which is equal to- 3.4322. The 
second method often avoids confusion and is 
usually to be preferred. 



130 PRACTICAL MATHEMATICS 

108. Powers. The power of a number is found by multiplying 
the logarithm of the number by the exponent. The number corre- 
sponding to this new log. is the desired power. 

Examples. 1. Find the fourth power of 24. 



Log. 2 


!4 = 


: 1.3802 

4 

5.5208 


Number = 


= 331800 


Log. 


23= 


= 1.3617 
5 


Number 


= i 


6.8085 
5434000 


= 9.9426- 


10 

7 





2. Find the fifth power of 23. 



3. "Find the seventh power of 0.8762. 

Log. 0.8762 

69.5982-70=1.5982 

Number= .3965 

109. Roots. The root of a number is found by dividing the 
logarithm of the number by the index of the root. The number 
corresponding to this new log. is the desired root. 

Examples. 1. Find the cube root of 8,753. 

Solution. The log. of 8,753 is 3.9421 Log. 8753 = 3.9421 

and dividing this by 3 gives 1.3140, which is 3)3.9421 

the log of 20.6, the desired cube root. Log. 20.6 = 1.31403 

2. Find the square root of 59.5. 

Solution. The division by 2 is carried Log. 59.5 = 1.7745 

out as if no decimal point was in the log., giv- 2)1.7745 

ing .8873, which is the log. of 7.713, the re- Log. 7.713 = .8873 
quired root. 

3. Find the sixth root of .9873. 

Solution. To avoid the minus char- Log. .9873 = — 1.9944 

acteristic, add and subtract some multiple — 1.9944 = 29.9944 — 30 

of 10 which will contain 6 a whole number 6)29.994 — 30 

of times. Therefore, adding and subtract- 4.9990 — 5 

ing 30 and dividing by 6 gives - 1.9990, Log. .996 = -1.9990 
which is the logarithm of .996. 



4. Solve 1/6539 

Log. 6539=3.8155-5-2=1.9077 
Number =80.85 



PRACTICAL MATHEMATICS 131 

5. Solve \3.65 

Log. 3.65 = 0.5G23 4-7- 0.0S03 
Number = 1.203 

6. Solve \i3o7) 5 

(367) 5 = 5xLog. 367 = 5X2.5047 = 12.8235 
But we must take the fourth root of (367) 5 and hence must divide 
its logarithm by 4. 

12.82354-4=3.2059. Number =1607 



4, 



Solve 



\ 



(1278)' 



(First calculate the quantity under the root sign.) 



8xLog. 1278=3.1066X8=24.8528 
4 X Log. 7=0.8451X4= 3.3804 

21.4724 
In order to extract the 6th root it is necessary to divide by 6, 
obtaining 3.5787, for which the number is 3791. 

8. Solve 4722 X^68 

Log. 4722= 3.6741 

+i Log. 68= 0.6108 

4.2849 

Number- 19270 

Special care must be taken in problems like the following: 
(231X244) -(113XL196) 

Subtraction can not be simplified by logarithms, therefore, only 
two quantities, the minuend and subtrahend, can be calculated 
separately by means of logarithms, finding their respective numbers, 
and subtracting one from the other. Thus: 

Log. 231 = 2.3636 
Log. 244 =2.3874 

4.7510 Number- 56360 

Log. 113=2.0531 
+i Log. 196 =0.7641 

2.8172 Number- 656.4 

Difference .55703.6 



L32 PRACTICAL MATHEMATICS 

Rules, (a) The logarithm of a power of 10, positive or negative, 
is a whole number, the mantissa being zero in every case. 

(b) // the logarithm is negative — true for every number betiveen 
and 1 — it is customary to add 10 or some multiple of 10 to the charac- 
teristic and to indicate a subtraction of the same number from the man- 
tissa. Thus, -3.7583 becomes 7.7583 - 10; - 13.8625 becomes 
7.8625-20. 

(c) If a number is greater than 1 , the characteristic of its logarithm 
is one unit less than the number of figures to the left of the decimal point. 

(d) // the number is less than 1, the characteristic of its logarithm 
is negative and one unit more than the number of zeros between the decimal 
point and the first significant figure of the given number. 

(e) Conversely, if the characteristic of a given logarithm is posi- 
tive, make the number of figures in the integral part of the corresponding 
number one more than the number of units in the characteristic. 

(f) If the characteristic is negative, make the number of zeros 
betiveen the decimal point and the first significant figure of the corres- 
ponding number one less than the number of units in the characteristic. 

(g) The logarithms of all numbers which differ from each other by 
powers of 10 have the same mantissa. 

PROBLEMS FOR PRACTICE 

Find the logarithms of the following: 

1. 53 4. 425 

2. 1,068 5. 650 

3. 821 6. 897 Ans. 2.9528 

Find the numbers corresponding to the following logarithms: 

7. 8.7284-10 Ans. .0535 10. 7.5862-10 Ans. .003856 

8. 1.8136 11. 0.8744 

9. 6.9956 12. 0.0085 Ans. 1.02 

Solve the following problems by logarithms : 

13. V826 Ans. 28.74 ^ «l<1021£ Ang 15Q5 

X (6) 3 
14 V^ 17. 5,026Xi/40 

15. ^(381? Ans. 35.36 i 8 . (1.030) 1 x(~) 



PRACTICAL MATHEMATICS 133 

CURVE PLOTTING 

110, Scope and Use of Curves. Graphical representation is the 
"shorthand" of the engineer, architect, and machinist. Everyone 
is more or less familiar with the detail drawings which the archi- 
tect and the contractor find so useful in showing the exact location 
of the different parts of a building. The machinist, too, easily 
works up a complicated piece of mechanism with only plans, eleva- 
tions, and sections to follow. There is another graphical method 
of showing the true relation of quantities, which is not so well 
known but which is none the less important, namely, curves. Like 
mechanical drawings, curves must be studied and the principles 
used in their construction must be learned before the proper mean- 
ing can be obtained from them, but with a little practice they become 
extremely useful not only in showing how one quantity changes 
with another, but in furnishing at a moment's notice information 
which would require hours of calculation to produce. 

The range of application is unlimited. For example, curves 
can be plotted showing the cost of doing contract work or of operat- 
ing a certain machine; the efficiency of a gang of men; the variations 
of temperature in a fever patient in a hospital ward; or the cost 
of high living. So accurate can these curves be drawn that the 
designer can use them in determining the allowable stress in struc- 
tural steel girders, or the contractor at a glance can tell the cost 
per foot for any diameter of sewer pipe placed at any depth. Thus 
the student can readily see that curves are valuable records for 
commercial and professional men, for trained engineers and skilled 
machinists. The use of curves and their meaning and the methods 
of constructing them are clearly shown in the following illustrations. 

PLOTTING METHODS 

111. Coordinate Paper. The curve in Fig. 55 has been drawn 
upon paper provided with parallel ruled lines; the lines drawn in 
both the horizontal and the vertical directions are equal distances 
apart, thus dividing the paper into small squares. The paper is 
called cross-section, or co-ordinate, paper and can either be ruled 
by the student or purchased already ruled at any stationer's store.* 

*A limited supply of co-ordinate paper will be supplied by the School when needed, 
and students are strongly recommend d r-, use it in practice work. Additional sheets will 
be sent on application at 10 cents per dozen sheets. 



134 PRACTICAL MATHEMATICS 

For convenience in plotting curves the side of each small square is 
one-tenth of an inch. 

1 12. Relation of Coordinates to Curve. After selecting co-or- 
dinate paper the first step in plotting curves consists in drawing a 
vertical line near the left edge of the paper and a horizontal line 
near the bottom. These two lines, shown as A and B in Fig. 55, 
are called axes and intersect in a point called the origin — some- 
times called the zero point. The vertical line is called the vertical 
axis and the horizontal line the horizontal axis. 

Referring to Fig. 55, it can be seen that the series of points P lf 
P2) P3 — Ps lie upon the curve, and, further, the positions of these 
points, since they lie upon the curve, determine the shape of the 
curve. If this curve remains in the same relative position as regards 
the horizontal and the vertical axes, each point of this curve must 
have a definite position in relation to these axes. For example, 
the point P 2 , to fall upon the curve, must be located 32 spaces 
up from the horizontal axis, and 10 spaces out to the right of the 
vertical axis. Again, point P 7 must be measured 20 spaces up from 
the horizontal axis and 35 spaces to the right of the vertical axis. 

If we supplied other values than these given for P 2 and P 7 , the 
student can readily see that the points P 2 and P 7 would not fall 
upon the curve but at some point either to the right or to the left 
of it. If this is so for two points, it will be so for any point on this 
curve or on any other curve; therefore, for any point on any curve, 
two values, one horizontal and one vertical, are necessary to locate 
a point, and these values, called the co-ordinates of the point, can only 
be used for locating this particular point. The distance measured 
on the vertical axis is called the vertical value and the distance on the 
horizontal axis the horizontal value. 

The term "plotting the curve" is, therefore, understood to 
mean the locating of the points of any curve by the use of co-ordi- 
nates of these points. 

Co-Ordinate Values. The small spaces, as measured on the 
vertical and on the horizontal axis, are usually drawn to a certain 
scale but may represent any values whatsoever. For example, one 
space on either axis may represent one foot, one yard, one year, 
one dollar, or any value that we choose to give it. If days be plotted 
or laid off on the horizontal axis and one day be represented by one 



PRACTICAL MATHEMATICS 



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Fig. 55. Textbook Study Record Curve 



Textbooks 


Study Hours 


1 


40 


2 


32 


3 


28.5 


4 


25.5 


5 


23.5 


6 


21.5 


7 


20. 


8 


18.5 



136 PRACTICAL MATHEMATICS 

space to the right of the origin, ten days are truly represented by 
ten of these spaces to the right of the origin, and so on for any num- 
ber of days. Again, it may be stated that one space equals ten years, 
in which case ten spaces will represent 10X10, or 100 years. 

113. Plotting the Curve. Study Record Curve It was shown 
that each point on the curve must have for its location two values, 
viz, one on the horizontal and the other 
on the vertical axis. Suppose, for 
example, a student wished to keep a 
record of his study hours spent on the 
first eight textbooks of his course. The 
two quantities to be studied by this 
curve are hours and textbooks. After 
looking over his study record card, he finds that he completed the 
first textbook in 40 study hours, the second in 32 hours, the third 
in 28.5 hours, the fourth in 25.5 hours, and so on as given above. 

To plot this curve, lay off the hour values on the vertical axis, 
Fig. 55, letting each space represent one hour; then for 40 hours 
count 40 spaces from the origin on the vertical axis. It is a good 
plan to write 10 hours, 20 hours, etc., just to the left of the vertical 
axis, as shown in the figure, to indicate how far up to go in order 
to get a certain number of hours without actually counting each 
space. Similarly, on the horizontal axis lay off the number of text- 
books to a scale of one textbook for every five spaces, writing the 
number just under the horizontal axis, as in Fig. 55. If the study 
card, or the table shows that it requires 40 hours of study on text- 
book No. 1, these values represent the co-ordinates of P\. Plot 
this point by reading from the origin to the right on the horizontal 
axis as far as textbook No. 1, then reading up the vertical line pass- 
ing through this point as far as the horizontal line representing 40 
hours; the intersection of the vertical and the horizontal lines is 
Pi, the point on the curve. Make a firm dot, mark it P lt and go 
to the next set of values. The table shows that the second text- 
book was completed in 32 hours. Read out 10 spaces to the right 
of the origin on the horizontal axis and up the vertical line to the 
horizontal line representing 32 hours; place a dot at the intersec- 
tion of these two lines, thus locating point P 2 . Continue in exactly 
the same manner until all points have been plotted for the eight 



PRACTICAL MATHEMATICS 



137 



textbooks. After all points have been plotted, draw a smooth line 
through the points and obtain the curve C. If the student is care- 
ful, he can draw in this curve freehand with a lead pencil, but in 
order to make a smooth line, an irregular curve of the type shown 
in Fig. 56 may be used to advantage. These curves can be purchased 
at any stationery store for a small sum. It is well to have one curve 
as well as a good ruler for this work. 

Transmission Rope Curve. To further illustrate the plotting 
materials let us take another practical case. The engineering 




Fig. 56. Typical Irregular Curves. These are made of Wood, Hard Rubber, or Celluloid 



department of a certain firm manufacturing transmission steel rope, 
having made tests of the horsepower developed by rope running at 
certain velocities, expressed this relation in the form of a curve. 
The velocities of the rope and the corresponding horsepowers devel- 
oped are arranged in tabular form above the curve, Fig. 57. 

The velocity of the rope in feet per second is chosen as the verti- 
cal value and the corresponding horsepower as the horizontal value, 
using a scale of 1 space on the vertical axis to equal 2 feet per second, 
and 1 space on the horizontal axis to equal 4 horsepower. To plot 
point Pj, lay off to scale on the horizontal axis 27 horsepower, remem- 
bering that one square equals 4 horsepower; that is, 27 horsepower 
will equal -V, or 6.75, spaces. For the vertical value, since each 



L38 



PRACTICAL MATHEMATICS 



space equals 2 feet per second, 10 feet per second equals -V 0- , or 5, 
spaces. Therefore, to plot point 1\, read to the right on the horizon- 
tal axis Of spaces and from this point up in a vertical line five spaces, 
placing a dot at this point to mark P x on the curve. Again, to 
plot point P 2 , read to right from origin - 5 ^, or 13.2, spaces and up 
on the vertical line at this point % -, or 10, spaces. Continue in the 
same manner until all the points are plotted. For the convenience 



100 

90 

eo 

70 
60 
50 
40 
30 

20 



1 1 1 II 1 1 1 1 1 1 1 1 1 1 ■""" ~~*p ' 








10 27 _ *_ ._ 


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Fig. 57. Horsepower Curve for Transmission Steel Rope 

of plotting and reading values, it is sometimes a good policy, before 
plotting any of the points, to lay off on the vertical axis all of the 
vertical values; that is, in this case 10 feet per second, 20 feet per 
second, 30 feet per second, etc. Then lay off to scale on the hori- 
zontal axis the horsepower values, as 27, 53, 79, etc., until all val- 
ues up to 247 have been located. After the vertical and the hori- 
zontal values have all been marked, it is a very simple matter to 
establish their intersections and thus locate the points. This curve 
will be found practically a straight line up to point P 5 , from which 
point the curve, although still a straight line, has a slightly steeper 
slope. Such a curve can be drawn with a straight edge or rule. 



PRACTICAL MATHEMATICS 139 

The student will note in this case and elsewhere that the term curve 
refers to the line drawn through the points whether such line proves 
to be a straight line or series of straight lines, or a curved line. Do 
not be confused, therefore, at the shape of your curve, and think 
because it is a straight line you have the wrong result. In most 
problems you will find, instead of regular curves or straight lines, 
the curve is made up of a series of straight lines between the points 
as shown in the Sunday load diagram, Fig. 65. 

MEANING OF CURVES 

114. Study Curve. The student will at first see no use for 
curves and, therefore, he should make a careful study of the curve, 
Fig. oo. First note that the curve drops gradually from -Pi to P 8 . 
The question arises, ""What does this mean?" This question can 
best be answered by asking another one. What sort of curve would 
have been obtained if the student had taken 40 hours for each text- 
book? Evidently such a curve would have been a straight line 
running along the 40 mark parallel with the horizontal axis. The 
drop in the curve towards the horizontal axis indicates that the stu- 
dent is taking less time to complete each succeeding text, the figures 
showing 40 hours for the first and 18.5 hours for the last. In other 
words, this curve is a graphical picture of the student's efficiency 
since he began to study. The completion of each text has contrib- 
uted its share to his brain development and added to his knowledge 
of the subject, making it easier for him to grasp the subject matter 
of his course with each completed text. Further, it shows at a 
glance how many hours he required to finish any one text. 

115. Horsepower Curve. The curve, Fig. 57, shows that, 
as the velocity of the rope in feet per second increases, the horse- 
power increases in the same proportion up to a point between points 
5 and 6. From this point to point 9 the curve rises at a greater 
angle, which shows that less horsepower is produced for the same 
increase in velocity of rope. The fact that the curve continues as a 
straight fine shows that the horsepower is still proportional to the 
velocity of the rope, but that this proportion is slightly different 
from that of the first part of the curve. 

The student has now seen that each curve is a graphical pic- 



140 



PRACTICAL MATHEMATICS 



ture of the relation of certain values which make up the curve. In 
other words, it is easier to use the curve than to wade through the 
mass of data making up the curve in order to obtain certain infor- 
mation regarding the action of one value on another. 



1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 




1 II II 1 1 II II II II 1 








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Fig. 5S. Curve of Squares and Square Roots 



ILLUSTRATIVE CURVES 

1. Curves for Squares and Square Roots. In Fig. 58 is shown 
a curve expressing the relation of numbers and the squares of those 
numbers. Curves of this kind are very practical, and, as can be 



PRACTICAL MATHEMATICS 



141 



readily seen, may be used to save time in all kinds of mathematical 
work. Compare, for example, the time required to find the square 
of 9.8 mathematically and the time to take the result directly from 
the above curve. To understand thoroughly the method of using 
such a curve, the following explanation will be helpful. Five divi- 
sions on the horizontal axis equal one unit and, therefore, each divi- 



to 



PUL LEY VIA M. IN INCHES 
ZO 30 40 SO 60 



70 80 



10 



20 



30 



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3 

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Fig. 59. Curve for Calculating Belt Lengths 



sion equals j, or tV, or .2. On the vertical axis there are five divisions 
to ten numbers and, therefore, each division equals j of 10, or 2. 
First find on the horizontal axis the point whose value is 9.8. It will 
be between 9 and 10, just one line short of division 10. Follow 
this line upward until it meets the curve and from this point go 
horizontally to the left until the vertical axis is reached. On this 
line read the value. It is between 90 and 100, three divisions above 
90. Each division represents 2, therefore three divisions equals G. 
This must be added to 90, giving as a final result 9G approximately. 



142 . PRACTICAL MATHEMATICS 

To find the square root of a number the above method would be 
reversed, using the same curve. Find the number on the vertical 
axis whose root is to be extracted, then follow horizontally across the 
sheet to the point of intersection with the curve and then downward 
to the horizontal axis. The value found there will be the square 
root; for instance, the square root of 96=9.8 approximately. 

2. Curve for Calculating Length of Belt. A very useful curve, 
Fig. 59, may be plotted by which the length of belt required for any 
sized pulleys between two shafts at any distance can be found at a 
moment's notice. Lay off pulley diameters in inches on both hori- 
zontal and vertical axes to the same scale and draw a diagonal line 
from the origin at an angle of 45 degrees as shown. Dividing this 
line into exactly twenty-one equal parts and drawing lines at right 
angles to the diagonal which meet the vertical and horizontal axes, 
the figure is completed. The method of using this curve is as fol- 
lows: Assuming 30-inch and 50-inch pulleys, take, say, 30 inches as 
the vertical value and 50 inches as the horizontal and find their point 
of intersection. This point falls about half way between divisions 
10 and 11. The length of belt wrapping the pulleys is, therefore, 
10.5 feet in length. Assuming 20 feet as the distance between 
shafts, the total length of belt will be twice this distance plus the 
length wrapping pulleys as found above, or (2X20) + 10.5 =50.5 feet. 

In engineering work there are many practical curves that can 
be made in a similar manner to this belt curve which will be great 
labor-saving devices where similar computations are being made 
day after day. It will readily be seen how much shorter time is 
required to find the length of pulley from the curve than if the 
mathematical formula was used. By the regular formula 

£=3.1416^±^+2Z 
2 

where L is length of belt in inches; /, distance between shafts in 
inches; D, diameter of large pulley in inches; and d, diameter of 
small pulley in inches. Substituting the same values as used above, 
that is, 30 inches and 50 inches, 

L=3.1416 ($2±?°) + (2 X 20) 12 

Li 

— 605.6 inches, or 50 feet 6 inches approximately. 



PRACTICAL MATHEMATICS 



143 



3. Gross Income Curve for Steam Raiiroad. A certain steam 
railroad carrying suburban traffic was opened for business in 189S 
and received an increasing amount of business up to the year 1911 
when there was a drop. The manager of the road was requested 
by the board of directors to present a record showing the gross 
income of the road for each year of operation. He presented the 
curve, Fig. 60. The co-ordinates of the points are given in connec- 
tion with the curve. Years are laid off on the horizontal axis and 
incomes on the vertical axis, 5 spaces on the former equaling 1 year 
and 1 space on the latter equaling $100,000. The first point is 
located as in the other cases by reading to the right of the origin on 





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1904 
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1907 
1908 
1909 
/9IO 
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1 ,30 0,00 

2 .300.000 
2,600.000 
3,200.000 

3 ,500.000 
3,800.000 
3.800.OO0 
3.900-000 
4,000.000 

4 .1 0.0 
4 ,200.000 
4.400.000 
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1902 1903 



1908 1909. 1910 



Fig. 60. Steam Railroad Gross Income Curve 

the horizontal axis one year; then reading up the vertical line to 
1500,000; the intersection of these two lines gives the point wanted 
on the curve. Continuing in like manner for the other points, 
lines are drawn through the points as noted in Fig. 60. The curve 
then shows at a glance that the income per year increased each year 
up to 1911 and here it dropped to 82,000,000. This drop was readily 
traceable to the influence of a parallel electric line. As this new road 
made regular trips throughout the day, and much more frequently 
than the steam road, the electric road received most of the suburban 
business. Tlii- curve shows the relative value of the yearly income 
much more clearly and strikingly than many pages of statistics. 



144 



PRACTICAL MATHEMATICS 



4. Contract Price for Sewer Pipe. A very interesting and 
practical example of the use of curves in Civil Engineering estimat- 
ing is shown in Fig. 61. Making use of the data obtained from a 



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TABLE OF 
CO-ORDINA TES 
























































































PRICE 


DEPTH 


























































































.00 
■ 95. 

7.42 

2.10 

2.775 


5 FT. 

8 » 
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20 " 
25 " 














































































































































































































































































































































































































































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5 10 15 20 25 

Fig. 61. Curve for Contract Price of Twelve-Inch Sewer Pipe 

large number of estimates of sewer work, an engineer has arranged 
the values in table form and drawn the curve as shown. By means 
of this curve, the estimator can tell at a glance without further cal- 



PRACTICAL MATHEMATICS 



145 



dilation what will be the contract price per foot to bid on 12-inch 
sewer pipe. 

The student should plot this curve from the co-ordinates given 
in the table, taking for the vertical value the price per linear foot 



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MONTH 


INCHES 
RAIN FALL 




































































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JAN. 
FEB. 
MAR. 
APR. 
MAY 
JUNE 
JULY 
AUG. 
SEPT. 
OCT. 
NOV. 
DEC. 


3.08 
230 
232 
2.72 
333 
3.42 
332 
3.02 
3.06 
2.43 
2.52 
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JAN FEB, MAR-, APR MAY JUNE JULY AUG. SEPT. OCT. NOV. DEC. 
Fig. 62. Mean Rainfall Curve 

of this 12-inch sewer and for the horizontal value the depth of excava- 
tion of the sewer. This curve applies only to a vitrified clay tile 

sewer 12 inches in diameter, but 
in practice curves could easily be 
drawn for every possible case. 

5. Mean Precipitation Curve. 
The student will notice in this 
curve, Fig. 62, that the months 
of the year are plotted to inches 
rainfall for each month. On the 
horizontal axis lay off the months 
to the right beginning with Janu- 
ary" at the origin, and letting one 
month equal five spaces. On the 
vertical axis lay off the rainfall 
in inches, using a scale of ten spaces for every inch. The curve 
shows at a glance that the precipitation is least in January and great- 



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80 
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BO 40 60 80 

REV. PER MIN. OF PROPELLER 

Fig. 63. Speed Curve for Steamboat 



100 



14G 



PRACTICAL MATHEMATICS 



est from May to July. The student will also notice that the rain- 
fall increases almost gradually from January to April and from this 
point jumps to a high value in May, remaining about the same 



6.00 



5.50 



5.00 



4.50 



4.00 



330 



3.00 



2.50 



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1.50 



1.00 



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PLR 

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fl YEAH 
















_ /#74 


05.93 
.42 
.95 
2.48 
.70 
1.22 
2.26 
1.73 
1.02 
2.32 
1.54 
3.06 
2.15 
2.46 
1.70 
2.15 
1.90 
2.66 
1.27 
2.52 
2.49 
2.19 
1.39 
1.53 
1.70 
2.79 
1.30 
2.46 
2.29 
t.66 
J. 56 
1.70 
2.01 
1 93 
/.86 
1.43 
2.24 








_ /5 75 








1 876 








1977 








7878 








1 879 








1880 








1881 








1882 








1883 








1884 








IS 85 








1886 








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1889 








1890 




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1891 




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1880 1890 1900 ■ 1910 

Fig. 64. Curve for Fire Loss per Capita 



through June and July, and gradually dropping to its lowest value 
again in January. 

G. Speed Curve for Steamboat. The curve shown in Fig. 63 
is an actual plot of the data found in the speed trial of a steamboat. 
A speeder-counter was used to obtain the number of revolutions 
per minute of the propeller shaft; the log gave the speed in knots 



PRACTICAL MATHEMATICS 



147 



per hour. Using revolutions of the shaft as horizontal values and 
knots as vertical values, the curve is plotted. Having the curve 
plotted for a limited range of speeds, the engineer could easily extend 
it and be able to take values for the speed at any given number of 
revolutions. 

7. Curve Showing Fire Loss per Capita. Statistical curves 
of the type in Fig. G4 show how much more easily one can grasp the 
situation from a curve than from a column of figures. It is neces- 
sary to study the figures up and down to get a comparative idea, 
while the curve shows at a glance when the loss was great and when 
small. Other similar curves could easily be made up covering all 
sorts of tabulated data. 

8. Sunday Current Load Curve. The current values in 
amperes delivered by the generators of a trolley system have been 
recorded in the power house for every period of the day by a proper 
recording instrument. These two quantities, time and amperes 
of current, being known, a curve may 
now be plotted showing the variations 
of current consumption throughout the 
day. This curve, Fig. 65, is an 
exact copy of one which is on the 
desk of the manager every morning. 
The managers of these large com- 
panies need to have their informa- 
tion systematized and every bit of 
data as to the operation of the road must be boiled down to its simplest 
form. This curve shows at a glance at what time the maximum 
load came on the line, the value of the load, and the variations dur- 
ing the day. Without this curve it would require, perhaps, hours 
for him to get the information from the data, while with the curve 
a five minutes' study will give him a clear conception of the whole 
situation. 

The student can go deeper into the meaning of this curve and 
find the reason for the heavy load between one o'clock in the after- 
noon and midnight. 

9. Locomotive Repair Curve. Sometimes it happens that for 
sake of comparison two curves are plotted on the same sheet. In 
the case of the curve A, Fig. 66, there is a reason for the drop in 



Time 


Load in 
Amperes 


12:00 p.m. 


37,500 


4:00 a.m. 
9:00 a.m. 
1:00 p.m. 


12,500 
28,000 
45,000 


3:00 p.m. 
4:00 p.m. 


78,000 
75,000 


5:00 p.m. 


80,000 


6:00 p.m. 

8:00 p.m. 

12:00 p.m. 


74,000 
78,500 
48,000 



US 



PRACTICAL MATHEMATICS 



the annual cost of repairs 
per locomotive, which can 
be made clear by the use 
of curve B, showing the 
increased efficiency of the 
employes in the shop in 
connection with this work. 
By plotting both curves on one sheet, one can see at a glance that 
the cost of repairs per locomotive falls as the efficiency of labor 
increases. The time in years over which the costs and efficiencies 
were taken is plotted on the horizontal axis. For the sake of clear- 
ness in reading the curves as well as plotting them, the vertical 





Per Cent 


Annual Repair 


Year 


Efficiency 


Cost for 




of Labor 


Locomotive 


1905 


55 


$2308 


1906 


82 


1677 


1907 


92 


1563 


1908 


87 


1683 


1909 


93 


1243 









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'i-L 



3 4 5 6 7 
FORENOON 



1011121 

NOON 



5 6 7 8 9 10 II 12 
AFTERNOON *f 



Fig. 65. Sunday Current Consumption Curve for an Electric Road 

values for the efficiency curve are indicated on the left and for the 
cost curve on the right. This system will again be used in the 
Water and Load diagram, Fig. 67. 

10. Water Power and Load Curves. In Fig. 67 are shown 
plotted on the same sheet two curves taken from the power record 
of a hydro-electric plant. Curve A is called the "head" curve, the 
term head, referring to the depth of water available to run the water 
turbines for the generation of power. The kilowatt curve shows 
the number of kilowatts generated at any time during the day or 



PRACTICAL MATHEMATICS 



149 



100 




2200 



2000 



1800 



1600 



1400 



1200 



1905 



1906 IS 07 /go8 

HORiZO/iTAL VALUES FOR CURVES "A"AND"B' 



1909 



Fig. 66. Locomotive Repair Curves, Showing Cost of Repairs and Efficiency of Labor 











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M/"» / / h « /Vx 


/5 ^ r/ ' / i / = ^s^ 


i / \/ \J % 


1 / 1 . . 1 


f " KILOWATT C'S<? YE B " 


^ 




\ / ^ — K ~^v /—^ 


\ — ^ ^-^ 






500 I i ! | | j """'. " 








,. ... 1 



13.00 



12.50 



12.00 



11.50 



k» 

11.00 k 

k 



10.50^ 

8 



70.00 



k 
9.00 



8.50 



12 3 6 

FORENOON. 



9 12 

noon 



6 9 12 

AFTERNOQH 



8.0O 



Fig. 67. Power Curves for Hydro-Electric Plant 



150 



PRACTICAL MATHEMATICS 



night. The vertical values for the "head" curve are given at the 
right and for the kilowatt curve at the left. 

In rinding the meaning of these curves the first point to be 
noticed is the rise in the kilowatt curve for a drop in the head curve. 
This indicates that 
when a load comes 
on the line, a cer- 
tain amount of head 
must be used up in 
producing this elec- 
trical energy, that 
is, water power has 
been transformed 
into electrical en- 
ergy. The curves 
are valuable in that 
they show at a 
glance the load on 
the line for any 
hour of the day and 
the head of water 
for any period of 
the 24 hours. It 
will be seen that as 
the load increases 
or decreases the head changes in the opposite direction. 

11. Comparison Curves for Incandescent Lamps. The two 
curves shown in Fig. 68 represent actual tests of the efficiency of 
the old type carbon filament lamps and the new tungsten filament 
lamps and give a striking graphical picture of the improvement of 
the new over the old style lamp. Horizontal values represent hours 
burned, and vertical values candlepower of the lamps. Note that 
the candlepower scale has been started for convenience at 9, it 
being evident that if it started at the curves would be higher up 
but otherwise the same. 

The meaning of the curves is clear. The carbon lamp 
reached the "smashing point," as it is called, in 1,000 hours, whereas 
the tungsten lamp had only dropped in candlepower to 91 per cent 



Time 


A. 


M. 


P. 


M. 


Head 


Total 


Head 


Total 




in 

Feet 


K. W. 


in 

Feet 


K. W. 


12:00 


12.29 


840 






12:30 


12.48 


770 


11.05 


1700 


1:00 


12.42 


770 


10.95 


1770 


1:30 


12.41 


920 


10.95 


1790 


2:00 


12.43 


850 


10.83 


1790 


2:30 


12.46 


830 


10.68 


1740 


3:00 


12.43 


860 


10.65 


1730 


3:30 


12.41 


840 


10.65 


1720 


4:00 


12.41 


770 


10.65 


1740 


4:30 


12.58 


870 


10.66 


1730 


5:00 


12.63 


840 


10.70 


1730 


5:30 


12.36 


940 


10.70 


1760 


6:00 


11.94 


1470 


10.70 


1790 


6:30 


11.62 


1700 


10.70 


1700 


7:00 


11.50 


1660 


11.00 


1560 


7:30 


11.25 


1770 


11.13 


1380 


8:00 


11.15 


1770 


11.19 


1300 


8:30 


11.00 


1760 


11.19 


1310 


9:00 


11.05 


1770 


10.80 


1620 


9:30 


11.00 


1760 


10.66 


1590 


10:00 


11.75 


1330 


10.64 


1590 


10:30 


11.55 


1440 


10.93 


1460 


11:00 


11.05 


1800 


11.04 


1310 


11:30 


10.95 


1800 


11.63 


1270 


12:00 


11.00 


1770 


12.40 


880 



PRACTICAL MATHEMATICS 



151 



*m 



7 &» 



AA / tp 



S& 



14 

1 



BURNING TIME 
IN HOURS 



CANDLE POWER 
CARBON 



CAUDLE POWER 
TUNGSTEN 







-hm 





too 

200 

400 

600 

800 

1000 

I 200 

1400 

1600 

1800 

2000 

2200 



16 

16.5 

16.1 

I 5.3 

14.5 

13.6 

12.8 



I 6 

16 

16 

16 

1 5.8 

15-7 

15.5 

15.2 

14.9 

14.7 

I 4.65 

I 4.55 

14.55 



ZOO '400 600 800 I00O 1200 1400 1600 1800 2000 2200 

NUMBER OF HOURS BURNED 
Fig. 6S. Candlepower Curves for Carbon and Tungsten Lamps 



200 




1875 1880 



1890 



1910 



Fig. 69. Curve Showing Variations in Market Price of Wheat 



152 



PRACTICAL MATHEMATICS 



of its original value in 2,200 hours. Further, the tungsten lamp 
held a uniform candlepower practically for 600 hours, while the 
carbon lamp began to decline rapidly after 100 hours of burning. 

12. Curve of Wheat Prices. The price of wheat per bushel 
varies considerably from year to year, as may be seen from Fig. 09. 
A fairly good idea of the variation can be obtained by running down 
the column of figures in the table, but it is not nearly as easy to 
carry the maximum and minimum prices as to have them pictured 



ll lll i ll l 



H 



TABLE 



DIAM. INCH. 



CURVE -A- 
R. P. M. 




DIAM. OF EMERY WHE^t 



10000 



3000 



8000 



7000 



6000 



5000 



4000 



3000 



2000 



1000 



20 Id 16 14 12 10 8 6 4 2 

Fig. 70. Speed Curves for Emery Wheels and Pulleys 

graphically by a curve. The curve is plotted from the data given 
in the table and a glance will show that the maximum price is $2 
per bushel in 1888 and the minimum price 63fc. in 1894. The 
value of such curves in all business lines where prices of commodi- 
ties vary from time to time is readily seen. 

13. Speed Curves for Wheels or Pulleys. The curve, Fig. 
70, shows how. for a certain surface or rim speed, the revolutions 



PRACTICAL MATHEMATICS 153 

per minute of an emery wheel is obtained when its diameter is known. 
In this particular case curve A and curve B represent the usual 
surface speeds of the two grades of emery wheels. Similar curves 
may be drawn for the usual surface speed of any wheel or pulley 
and then with the diameter given the revolutions may be obtained, 
or with the revolutions given the diameter may be obtained. Such 
a curve is a very practical one and useful in all kinds of machine 
shcps. 

CURVE PROBLEMS FOR PRACTICE 

1. Plot the curve from the following data for the cost of high- 
speed compound engines, using thousands of dollars for vertical 
values and horsepower for horizontal values. What will be the 
cost of a 750 horsepower engine of this type? 

Dollars Horsepower 

6400 400 

9600 600 

12800 800 

16000 1000 

24000 1500 

32000 2000 

40000 2500 

48000 3000 

2. Plot the curve from the following data for the increase of 
Chicago's population, using population in thousands as vertical 
values and years for horizontal values. Assuming the same increase 
in the next decade as in the last, find the population of Chicago in 
1920. 

Population Year3 

109,206 1860 

178,492 1865 

306,605 1870 

400,500 1875 

503,185 1880 

665,000 1885 

1,099,850 1890 

1,366,813 1895 

1,698,575 1900 

1,949,116 1905 

2,195,551 1910 

3. Plot the curve from the following data for the areas and 
diameters of circles, using areas for vertical values and diameters 
for horizontal values. What is the diameter of a circle having an 
area of 40 square inches? 



154 PRACTICAL MATHEMATICS 

Diameter 
Inches 

1 

2 

3 

4 

5 

6 

7- 

8 

9 

10 
11 
12 

4. Plot the curve from the following data for the steam con- 
sumption in a steam turbine, using pounds of steam per h.p. hour 
as vertical values, and horsepowers as horizontal values. Find 
from the curve the steam consumption per h.p. hour of a 450-h.p. 
turbine. 





Area 


Sq. 


Inches 




.7854 


3 


.1416 


7 


.0686 


12 


. 566 


19 


.635 


28 


.274 


38.485 


50 


.265 


63 


.617 


78.540 


95 


033 


113 


097 



ounds Steam 


Horsepower 


61.75 


300 


49.25 


400 


40 


500 


34.75 


600 


34 


700 


38 


800 



5. Plot the curve from the following data for a centrifugal 
pump. Plot the values given in the table, using gallons per minute 
as vertical values, and horsepower per foot of lift as horizontal 
values. Find from the curve how many gallons of water can be 
lifted per foot per minute with a horsepower of 25. 

Gals, per Min. p^FoTuft - 

10.000 4.1 

20^00 8. 

40,000 15.6 

80,000 31.4 

100,000 39. 

120,000 46.75 

130,000 51. 



6. Plot a smooth curve showing the water consumption 
hour for a 400-kilowatt steam turbine by means of the following 
values. Plot pounds of water per hour as vertical values and horse- 
power as horizontal values. From the curve determine how many 
pounds of water per hour will be required when the steam turbine is 
delivering 700 horsepower. 



per 



PRACTICAL MATHEMATICS 155 

Lbs. 



Water per Hr. 


Horsepower 


5,000 


300 


7,500 


500 


11,000 


750 


13,000 


875 


17,500 


1100 



7. Plot the following values for load and efficiency for an 
air-blast transformer and draw a smooth curve through points 
plotted. Plot per cent efficiency as vertical values and per cent 
loads as horizontal values. Determine from the curve at what 
load the highest efficiency of the transformer may be expected. 

Per Cent Load 

150 

125 

100 

75 

50 

25 

10 

5 

2.4 
1.25 




Per Cent 


Efficiency 


98.20 


98.29 


98.32 


98.24 


97.88 


96.43 


92.00 


85.00 


70.00 


50.00 






156 PRACTICAL MATHEMATICS 

APPENDIX 

MEASURES OF EXTENSION 
Linear Measure 

12 inches (in. or ") = 1 foot (ft. or ') 

3 feet = 1 yard (yd.) 

5 h yards or 16£ft. = 1 rod (rd.), pole, or perch 

320 rods or 5280 ft. = 1 mile (mi.) 

A knot, used in navigation, is equal to 1.15 miles. 
A fathom is equal to six feet. 

Square Measure 

144 square inches (sq. in.) = 1 square foot (sq. ft.) 

9 square feet = 1 square yard (sq. yd.) 

30^ square yards = 1 square rod (sq. rd.) 

160 square rods = 1 acre (A.) 

640 acres = 1 square mile (sq. mi.) 

Cubic Measure 

1728 cubic inches (cu. in.) = 1 cubic foot (cu. ft.) 

27 cubic feet = 1 cubic yard (cu. yd.) 

128 cubic feet = 1 cord (cd.) 

8 cord feet = 1 cord (cd.) 

A 'perch of stone or masonry is 16 J feet (1 rod) long, 1 J feet wide, 
and 1 foot high, or 24| cu. ft. capacity. The perch sometimes has 
other values, viz, 22 cu. ft. and 18 cu. ft. The value first mentioned 
is used throughout this text. 

16 cu. ft. = 1 cord foot (cd. ft.). A pile of wood 8 feet long, 4 
feet wide, and 4 feet high contains a cord. The cord foot is 1 foot 
of the length of such a pile. 

Surveyors' Linear Measure 

7.92 inches (in.) = 1 link (1.) 
25 links = 1 rod (rd.) 

100 links ] 

4 rods each = 1 chain (ch.) 

66 feet ) 
80 chains « 1 mile (mi.) 



PRACTICAL MATHEMATICS 157 

Surveyors' Square Measure 

16 square rods = 1 square chain 

10 square chains = 1 acre (A.) 

640 acres = 1 square mile (sq. mi.) 

1 square mile = 1 section (sec.) 

36 sections = 1 township (Tp.) 

Surveyors' measure was obtained by calling one tenth of an 
acre a square chain. A tenth of an acre is equal to 16 square 
rods, and is equivalent to a square each side of which measures 4 
rods. Thus, since 16 square rods is equal to one square chain, a 
linear chain is 4 rods or 792 inches. The linear chain is divided 
into 100 equal parts called links. 

MEASURES OF CAPACITY 
Dry Measure 

2 pints (pts.) = 1 quart (qt.) 
8 quarts = 1 peck (pk.) 

4 pecks = 1 bushel (bu.) 

The Winchester bushel contains 2,150.4 cubic inches and is used 
in measuring shelled grains. The heaped bushel of 2747.7 cubic 
inches is used for measuring apples, potatoes, corn in the ear, etc 
The dry gallon, or half peck, contains 268.8 cubic inches. 

Liquid Measure 

4 gills (gi.) = 1 pint (pt.) 

2 pints = 1 quart (qt.) 

4 quarts .= 1 gallon (gal.) 

31} gallons = 1 barrel (bbl.) 

2 barrels, or 63 gallons = 1 hogshead (hhd.) 

One U. S. standard gallon contains 231 cubic inches, and one 
gallon of water weighs approximately 8J pounds. 

One cubic foot of water contains approximately 7.5 U. S. standard 
gallons and weighs 62.3 pounds. 

Apothecaries' Fluid Measure 

60 minims = 1 fluid drachm 

8 fluid drachms . = 1 fluid ounce 
lo fluid ounces = 1 pint 



15S PRACTICAL MATHEMATICS 

MEASURES OF WEIGHT 
. Troy Weight 

24 grains (gr.) = 1 pennyweight (dwt. or pwt.) 
20 pennyweights = 1 ounce (oz.) 
12 ounces = 1 pound (lb.) 

The standard Troy pound contains 5,760 grains and is identical 
with the Troy pound of Great Britain. Troy weight is used in 
weighing gold, silver, and jewels. 

The carat (car.), a weight of about 3.2 grains Troy, is used in 
weighing diamonds and other precious stones. The term carat is 
used also to express the fineness of gold, and means 2 - 1 T part. For 
example, gold is said to be 18 carats fine when it contains 18 parts 
of pure gold and 6 parts of alloy or baser metal. 

Avoirdupois Weight 

16 ounces (oz.) = 1 pound (lb.) 

100 pounds (lbs.) = 1 hundredweight (cwt.) 

20 hundredweight, or 2,000 lbs. = 1 ton (t. or T.) 

The Avoirdupois pound contains 7,000 grains Troy, 
There are 7.3 Apothecary drams in an Avoirdupois ounce. 
25 pounds are sometimes called a quarter. 
The ton of 2,000 pounds is sometimes called the short ton. 

Long Ton Weight 

16 ounces = 1 pound (lb.) 

112 pounds (lbs.) = 1 hundredweight (cwt.) 
20 cwt. or 2,240 lbs = 1 ton (T. or t.) 
Note. The long ton table is not to be used unless specially designated. 

Apothecaries Weight 

20 grains = 1 scruple (sc.) 

3 scruples = 1 dram (dr.) 

8 drams = 1 ounce (oz.) 

12 ounces = 1 pound (lb.) 

The pound, ounce, and grain of this weight are identical with 
those of Troy weight. 

Drugs and medicines are bought and sold at wholesale by 
Avoirdupois weight. 



PRACTICAL MATHEMATICS 



159 



METRIC SYSTEM 

The fundamental unit of the metric system is the meter — the 
unit of length. From this the units of capacity — liter — and of weight 
— gram — were derived. All other units are the decimal sub-divisions 
or multiples of these. These three units are simply related; e. g., * 
for all practical purposes one cubic decimeter equals one liter and one 
liter of water weighs one kilogram. The metric tables are formed 
by combining the words meter, gram, and liter with the six numerical 
prefixes, as in the following tables: 



Prefixes 




Meaning 




Units 


milli 


= 


one thousandth 


ltfotf 


.001 




centi- 


= 


one hundredth 


i 

TTTD 


.01 


meter for length 


deci- 


= 


one tenth 


A 


.1 




Unit 


= 


one 




1 


gram for weight or 
mass 


deka- 


= 


ten 


¥ 


10 




hecto- 


= 


one hundred 


100 


100 


liter for capacity 


k ilo- 


= 


one thousand 


100 


1000 





Units of Length 


milli-meter = 




.001 


meter 


centi-meter = 




01 


meter 


deci-meter = 




1 


meter 


METER 


1 




meter 


deka-meter = 


10 




meter 


hecto-meter = 


100 




meter 


kilo-meter = 


1,000 




meter 



Where miles are used in England and the United States for 
measuring distances, the kilometer (1,000 meters) is used in metric 
countries. It is about 5 furlongs. There are about 1,600 meters in 
a statute mile, 20 meters in a chain, and 5 meters in a rod. 

The meter is used for dry goods, merchandise, engineering con- 
struction, building, and other purposes where the yard and foot are 
used. The meter is about a tenth longer than the yard. 

The centimeter and millimeter are used instead of the inch and its 
fractions in machine construction and similar work. The centimeter, 
as its name shows, is the hundredth of a meter. It is used in cabinet 



The term "e. g." means "for example. 



160 PRACTICAL MATHEMATICS 

work, in expressing sizes of paper, books, and in many cases where 
the inch is used. The centimeter is about two-fifths of an inch and 
the millimeter about one twenty-fifth of an inch. The millimeter 
is divided for finer work into tenths, hundredths, and thousandths. 
If a number of distances in millimeters, meters, and kilometers 
are to be added, reduction is unnecessary. They are added as dollars, 
dimes, and cents are now added. For example, "1,050.25 meters" 
is not read "1 kilometer, 5 dekameters, 2 decimeters, and 5 centi- 
meters/' but "one thousand fifty meters and twenty-five centimeters," 
just as "$1,050.25" is read "one thousand fifty dollars and twenty- 
five cents." 

Area 

The table of areas is formed by squaring the length measures, 
as in o v .r common system. For land measure 10 meters square is 
called an are (meaning area). The side of one are is about 33 feet. 
The hectare is 100 meters square, and, as its name indicates, is 100 
ares, or about 2\ acres. An acre is about 0.4 hectare. A standard 
United States quarter section contains almost exactly 64 hectares. A 
square kilometer contains 100 hectares. 

For smaller measures of surface the square meter is used. The 
square meter is about 20 per cent larger than the square yard. For 
still smaller surfaces the square centimeter is used. A square inch 
contains about 6J square centimeters. 

Volume 

The cubic measures are the cubes of the linear units. The 
cubic meter (sometimes called the stere, meaning solid) is the unit of 
volume. A cubic TTteter of water weighs a metric ton and is equal 
to 1 kiloliter. ' The cubic meter is used in place of the cubic yard and 
is about 30 per cent larger. This is used for "cuts and fills" in grading 
land, measuring timber, expressing contents of tanks and reservoirs, 
flow of rivers, dimensions of stone, tonnage of ships, and other places 
where the cubic yard and foot are used. The thousandth part of the 
cubic meter (1 cubic decimeter) is called the liter. (See table of 
capacity units.) 

For very small volumes the cubic centimeter (c.c. or cm 3 .) is used. 
This volume of water weighs a gram, which is the unit of weight or 
mass. There are about 16 cubic centimeters in a cubic inch. The 



PRACTICAL MATHEMATICS 161 

cubic centimeter is the unit of volume used by chemists as well as 
in pharmacy, medicine, surgery, and other technical work. One 
thousand cubic centimeters make 1 liter. 

Units of Capacity 



milli-liter 


.= 


.001 liter 


centi-liter 


= 


.01 liter 


deci-liter 


= 


. 1 liter 


*LITER 


= 


1 liter 


deka-liter 


= 


10 liter 


hecto-liter 


= 


100 liter 


kilo-liter 


= 


1,000 liter 



The hectoliter (100 liters) serves the same purpose as the United 
States bushel (2,150.42 cubic inches), and is equal to about 3 bushels, 
oi a barrel. A peck is about 9 liters. The, liter is used for measure- 
ments commonly given in the gallon, the liquid and dry quarts, a liter 
being 5 per cent larger than our liquid quart and 10 per cent smaller 
than the dry quart. A liter of water weighs exactly a kilogram, i. e., 
1,000 grams. A thousand liters of water weigh 1 metric ton. 

Units of Weight (or Mass) 

milli-gram = 0.001 gram 

centi-gram = .01 gram 

deci-gram = .1 gram 

GRAM = 1 gram 

deka-gram = 10 gram 

hectogram = 100 gram 

fkilo-gram = 1,000 gram 

Measurements commonly expressed in gross tons or short tons 
are stated in metric tons (1,000 kilograms). The metric ton comes 
between our long and short tons and serves the purpose of both. The 
kilogram and half kilo serve for everyday trade, the latter being 
10 per cent larger than the pound. The kilogram is approximately 
2.2 pounds. The gram and its multiples and divisions are used for 
the same purposes as ounces, pennyweights, drams, scruples, and 
grains. For foreign postage, 30 grams is the legal equivalent of the 
avoirdupois ounce. 

*One liter equals 1.05668 liquid quarts or 0,9081 dry quarts. 
tOne kilogram equals 2.204622 avoirdupois pounds. 



162 PRACTICAL MATHEMATICS 

MEASURES OF TIME 



60 seconds (sec.) = 1 minute (min.) 

60 minutes = 1 hour (hr.) 

24 hours = 1 day (d.) 

7 days = 1 week (w. or wk.) 

365 days or 52 wks. = 1 year (yr.) 
12 months (mos.) = 1 year 

366 days = 1 leap year 
100 years (yrs.) = 1 century 



In most business transactions 30 days are considered a month 
and 360 days a year. 

The solar year is exactly 365 d. 5 hrs. 48 min. 49.7 sec. 

A common year consists of 365 days for three successive years. 
Every fourth year, except as noted below, one day is added for the 
excess of the solar year over 365 days, and we have 366 days, which 
make what is called a leap year. The extra day is added to the month 
of February, which then has 29 days. 

The following rule for leap year will make the calendar correct 
to within one day for a period of 4,000 years. 

Every year exactly divisible by 4 ^ s & leap year, the centennial 
years excepted; the other years are common years. Every centennial 
year exactly divisible by JflO is a leap year; the other centennial years 
are common years. Thus 1904 is a leap year, but 1905 is a common 
year; also, the year 2000 is a leap year, but 1800 and 1900 are common 
years. 

MEASURES OF MONEY 
United States Money 



10 mills (mi.) = 1 cent (£) 

10 cents = 1 dime (d.) 

10 dimes = 1 dollar ($) 

10 dollars = 1 eagle 



These values, with the exception of the eagle, apply also to 
Canadian money. 



PRACTICAL MATHEMATICS 163 

English Money 

4 farthings (far.) = 1 penny (d) 
12 pence = 1 shilling (s) 

20 shillings = 1 pound (£) or sovereign 

The unit of English money is the pound sterling, the value of 
which in United States money is $4.8665. 

French Money German Money 

1 franc = SO. 193 100 pfennige = 1 mark (Rm.) = $0,238 

MISCELLANEOUS MEASURES 

Measures of Angles and Arcs - Stationers' Table 

60 seconds (") = 1 minute (') 24 sheets = 1 quire (qr.) 

60 minutes = 1 degree (°) 20 quires = 1 ream (rm.) 

90 degrees = 1 right angle or quadrant 2 reams = 1 bundle (bdl.) 

360 degrees = 1 circle 5 bundles = 1 bale (bl.) 

Comparative table of Metric and English equivalents.* 

Linear Measure Solid Measure 



1 inch 
1 foot 
1 meter 
1 yard 
1 meter 
1 rod 
1 mile 
1 kilo 



2.5 cm. 
.3 meter 
39.37 inches 
.9 meter 
yards 
meters 
kilometers, 
miles 



1.1 
5. 
1.6 
.62 



1 cu. inch 
1 cu. cm. 
1 cu. foot 
1 cu. dm. 
1 cu. yard 
1 cu. meter 
1 cord 



= 16.4 cu. 
.06 cu. 
= 28.3 cu. 
= .035 cu. 
= .76 cu. 
= 35.3 cu. 



cm. 

in. 

dm. 

feet 

meter 

feet 



3.6 stares 



Square Measure 



Measure of Capacity 



1 sq. 
1 sq. 
1 sq. 
1 sq. 



inch = 
cm. = 
foot = 
meter = 



1 sq. yard 
1 sq. rod 
1 acre 
1 are 



.45 sq. cm. 
.155 sq. in. 
.09 sq. meter 

sq 

sq 

sq 



10.73 
.84 
25.3 
40.5 



feet 

meter 

meters 



.024 acres 



1 fl. ounce 
1 liq. qt. 
1 liter 
1 gallon 
1 dry qt. 
1 liter 
1 bushel 
1 hectoliter 



.03 litar 

.95 liter 
1.05 liq. qt. 
3.8 liters 
1.1 liters 

.9 dry qts. 

.35 hectoliter 
2.84 bushels 



1 grain Troy = 
1 gram = 

1 ounce Troy = 
1 lb. Troy 
1 kilo 



.06 gram 
15.4 grains Troy 
31. grams 
.37 kilogram 
2.7 lbs. Troy 



Measures of Weight 

1 ounce Avoir. 

1 gram 

1 lb. Avoir. 

1 kilo 

1 metric ton 



= 28.35 grams 
= .035 oz. Avoir. 

= .45 kilogram 

2.2 lbs. Avoir. 
= 2200. lbs. Avoir. 



♦Values given in these tables are only approximate but are sufficiently accurate for all 
ordinary computations. These tables need not be memorized. 



164 PRACTICAL MATHEMATICS 

THE SLIDE RULE 

In its present perfected form the slide rule has become an indis- 
pensable aid not only to the engineer and the scientist, but also to 
the manufacturer, the merchant, the accountant, and all others 
whose occupation or business involves calculations. 

The slide rule is an instrument for working mathematical calcu- 
lations easily and rapidly, and consists of a grooved base F into 
which a slide E is fitted, as shown in the illustration. The rule is 
graduated in a series of four scales A, B, C, and D, and is provided 
with a runner R which can be moved back and forth to find coin- 
ciding points. The starting and ending graduations of each scale 
are marked 1, and are called the left and right indices. The gradua- 
tions A and B on the lower edge of the rule are called the lower scales, 
and C and D on the upper edge, the upper scales. A and B are usually 
used for problems in multiplication, division, and proportion, as their 
length gives more sub-divisions, and thus more accurate results. 

METHOD OF USING THE SLIDE RULE 

Proportion. To illustrate the use of the slide rule in solving a 
problem in proportion, first set the left-hand indices of A and B in 
coincidence; then each reading on B will bear the same ratio to the 
coinciding reading on A as the ratio between the indices, viz, a ratio 
of 1:1. Now if, by moving the slide to the right, 1 on B is made to 
coincide with 2 on A the readings on B will then have the ratio of jl :2 
to the coinciding readings on A. Therefore, to solve a proportion, 
use the following rule: 

Rule. Set the B scale so that the first term of any proportion on 
this scale is coincident with the second term on the A scale. Then, 
without moving the slide, find the third term of the proportion on the 
B scale, and the coincident number on the A scale will be the unknown 
fourth term. 

It is important to observe that the values which the several 
divisions of the scales may have will depend directly upon the value 
assigned to the left-hand index figure on the scale. On the slide rule 
shown, this index figure is designated 1. It may, however, be taken 
to represent any value that is a multiple or sub-multiple of 10. The 
left-hand index 1 of the A scale may be regarded as 1, 10, 100, 1,000, 



PRACTICAL MATHEMATICS 



etc., or as .1, .01, .001, .0001, etc. When the 
initial value has been assigned to the index, the 

ratio of value must then be maintained through- 
out the whole scale. Tor example, if 1 on A is 
taken to represent 10, the main divisions 2, 3, 4, 
etc., are read as 20, 30, 40, etc. On the other 
hand, if the fourth main division is read as .004, 
then the left index figure of the scale is read as 
.001. 

Examples. 1. Find the unknown term in 
the proportion 10 : 19 : : 2 : x. 

Solution. Take 1 (10) on B, and move the slide 
to the right until the 1 coincides with 1.9 (19) on A; 
then under 2 on B, the value 3.S is found on A, which 
i>. therefore, the unknown term of the proportion. If 
the index does not fall exactly on a graduation, an esti- 
mation must be made. 

2. Solve 6 :1S ::2 : .r. 

Solution. It will be noticed that the scales begin 
at the left with 1, and not with zero. Therefore, the 
reduction made in Example 1 is not possible here, as 
.6 6) cannot be found on B. If, however, 6 on B is 
brought into coincidence with 1.8 (IS) on A, it will be 
found that 2 on B is beyond the limits of A. Therefore, 
proceed as follows: With 6 on B coinciding with 1.8 on 
A, set the runner so that the hair is oA*er the right in- 
dex of B. Xow push the slide to the right until the 
left index comes under the hair. Then under 2 on B, 
the reading 6 will be found on A. 6 is, therefore, the 
fourth or unknown term of the proportion. In solving 
examples such as this, the slide may, of course, be 
moved to the right or to the left, as may be required 
according to the numbers involved. 

Multiplication. Rule. Move the slide so 
tli nt the index on the B scale coincides with the 
multiplicand on the A scale; then the product can 
(>< read on the .1 scale under the multiplier on the 
B scale. If preferred, the slide may be moved 
so that the index on the B scale coincides with 
the multiplier on the A scale; then the product 
can be read on the A scale under the multiplicand 
on the B scale. 




166 PRACTICAL MATHEMATICS 

Examples. 1. Multiply 25 by 15. 

Solution. Move the slide until the left index of B is over 2.5 on A. 
Then under L.5 on B, 3.75 will be found on A. The desired product is, there- 
fore, :>7f>, as it is clear by an inspection of the problem that the product must 
contain three digits. 

2. Multiply 361 by 119. 

Solution. Move the slide so that the index on B is over 3.61 on A. 
Under 1.19 on B, read 4.29 on A. It may be seen by inspection that the prod- 
uct will have five numerical places. The answer is, therefore, 42,900 approxi- 
mately. In large numbers, such as those used in this example, approximate 
results only can be obtained, but they are sufficiently accurate for all practical 
purposes. 

Division. Rule. Move the slide so that the divisor on the B 
scale will coincide with the dividend on the A scale. The index on 
the B sccde will then coincide with the quotient on the A scale. 

Examples. 1. Divide 256 by 16. 

Solution. Move the slide so that 1.6 on B coincides with 2.56 on A. 
Coinciding with the left-hand index of B, 1.6 will be found on A. The quotient 
is 16. 

2. Divide 962 by 13. 

Solution. Move the slide until 1.3 on B coincides with 9.62 on A. 
Coinciding with the left-hand index on B, 7.4 will be found on A. The quotient 
is 74. 

Square and Square Root. The upper scale D is so related to the 
lower scale A that readings taken on D are the squares of the coincid- 
ing readings on A; while the readings on A are, of course, the square 
roots of the coinciding readings on D. 

Examples. 1. Find the square of 25.6. 

Solution. Move the runner so that the hair coincides with 2.56 on A. 
The hair is then found to coincide with 6.55 approximately on D. The square 
is, therefore, 655. 

2. Extract the square root of 311. 

Solution. Move the runner to 3.11 on D. The hair will then be found 
to coincide with 1.76 on A. 17.6 is, therefore, the square root required. 

Cube. Rule. To find the cube of a number, move the slide until 
the right or left index of the B scale coincides with the given number 
as located on the A scale. Then over identically the same number on 
the C scale find the cube of the number on the D scale. 



PRACTICAL MATHEMATICS 167 

Example. Find the cube of 9.3. 

Solution'. Move the slide until the right index of B coincides with 
9.3 on .4. The reading 9.3 on C will then coincide with 8.04 on D. The 
desired cube is, therefore, 804. 

It is possible by means of the slide rule to obtain the cube root 
and higher roots of numbers, but the process is so involved that the 
use of logarithms is recommended. 

Trigonometric Problems. If the slide E is completely withdrawn, 
three scales will be found on the reverse side. S, the upper one, 
is a scale of natural sines; T, the lower one, is a scale of natural 
tangents, and L, the middle one, is a scale of equal parts giving the 
logarithms corresponding to the numbers on A. When sines or 
tangents of angles are desired place the slide in the groove with this 
side up, and with the right and left indices coinciding. 

Rules. The angles will be found on S and their corresponding 
sines ivill be found on D. Those found on the left half of D must be 
divided by 100; while those on the right half must be divided by 10. 
For example, sin 2° 30' = 0.0436; sin 35° 30' = 0.581. 

To find the tangent of any angle, find the angle on T. The tangent 
will then be the coincident reading on A, divided by 10. For example, 
tan 21° = 0.3S4. Tangents from 5° 43' to 45° only, are given on 
the scale. The tangent and sine are approximately equal for angles 
less than 5° 43'. For angles larger than 45°, use the formula 

1 



tan A 



tan (90° - A) 



Logarithms. The method of finding the logarithms of numbers 
is as follows: 

Rule. With the slide in its original position, move it so that the 
left index of B coincides with the number on A of which the logarithm 
is to be taken. Reversing the entire slide rule, the logarithms may be 
read on the middle scale L, the number coinciding with the index in 
the notch on the base of the ride. For example, to find the logarithm 
of 5, move the slide to the right until the index on B coincides with 
5 on A. Without moving the slide, reverse the slide rule and the 
reading on L is found to be 7. The logarithm is, therefore, .700. 

Inverting the Slide. When the slide is removed and turned end 
for end, so as to bring the B scale adjacent to the D scale, and is moved 



168 PRACTICAL MATHEMATICS 

until the indices coincide, the readings on B are the reciprocals of 
the coinciding readings on A, and vice versd. This is useful in 
inverse proportion. 

Inverse Proportion. When in a problem more requires less or 
less requires more, an inverse proportion must be formed. This 
proportion may easily be solved by inverting the slide, when it will 
be found that the coinciding divisions of the scales, A and B, form 
a series of inverse ratios. By means of the runner, the readings may 
be readily taken off. 

Examples. 1. If 12 carpenters build a house in 48 days, how 
long would it take 16 carpenters to build the same house? 

Solution. Move the slide inverted to the left until 1.2 on B coincides 
by means of the runner with 4.8 on A. Then 1.6 on B will be found, by means 
of the runner, to coincide with 3.6 on A. The 16 carpenters will take 36 
days to build the house. 

2. A train running 20 miles per hour covers a certain distance 
in 7 hours. How long will it take a train running at 35 miles per 
hour to cover the same distance? 

Solution. Move the slide to the right until 2.0 on B coincides with 7 
on A. Then 3.5 on B will coincide with 4 on A. The faster train will accom- 
plish the run in 4 hours. 

Location of Decimal Points. In all of the problems so far used 
in this article, the number of digits in the various products may be 
readily noticed by inspection, and the decimal point located. For 
example, multiplying 25 by 15 gives three digits in the result. If 
the number of digits in the result is not readily noticed by inspection, 
the following method will be found useful. 

Multiplication. When the slide projects to the left, the number 
of digits in a product equals the sum of the digits in the multiplier 
and multiplicand, and when the slide projects to the right, the number 
of digits in the product equals the sum of the digits in the multiplier 
and multiplicand less 1. 

Example. 1.5 X 17 = 25.5. 

Solution. The sum of the digits of the two factors is 3, and the slide 
projects to the right. The product will, therefore, contain 3 — 1 or 2 digits. 

Division. The number of digits in a quotient when the slide 
projects to the left equals the number of digits in the divisor sub- 
tracted from the number of digits in the dividend, and when the slide 



PRACTICAL MATHEMATICS 169 

projects to the right equals the number of digits in the divisor sub- 
tracted from the number of digits in the dividend plus 1. 
Example. 22.5 ~ 15 = 1.5. 

Solution. Subtracting the number of digits in the divisor from the 
number of digits in the dividend gives zero. Since the slide projects to the 
right, this difference is increased by 1, giving one digit in the quotient. 

A little practice with problems of this sort will make it a simple 
matter to correctly determine the position of the decimal point. 

A perusal of the article on the slide rule may have impressed the 
reader with the fact that it is a rather complicated instrument. Its 
value depends in a large measure upon the quickness with which 
the operator can assign the correct value to each graduation of the 
different scales. There are, of course, certain limitations to its use 
on account of the size of the scales, which limitations can only be 
discovered by repeated trials and the results in many cases, especially 
where the numbers are large, can only be approximated. It must, 
therefore, be left to the student himself to determine the extent to 
which he will develop an ability to use the instrument, bearing in 
mind that many engineers have found it exceedingly useful in their 
work. 



170 



PRACTICAL MATHEMATICS 





NATURAL SINES, COSINES, TANGENTS, 


AND COTANGENTS* 


Deg. 


Sin. 


Cos. 


Tan. 


Cot. 


Deg. 


Sin. 


Cos. 


Tan. 


Cot. 





. 00000 


1 . 0000 


. 00000 


00 


46 


.71934 


.69466 


1 . 0355 


. 96569 


1 


.0174.') 


. 99985 


.01746 


57.290 


47 


.73135 


.68200 


1.0724 


.93252 


2 


. 03490 


. 99939 


.03492 


28.636 


48 


.74314 


.66913 


1.1106 


. 90040 


3 


.05234 


. 99863 


.05241 


19.081 


49 


.75471 


.65606 


1 . 1504 


. 86929 


4 


. 06976 


.99756 


.06993 


14.301 


50 


.76604 


.64279 


1.1918 


.83910 


5 


.08716 


.99619 


.08749 


11.430 


51 


.77715 


.62932 


1.2349 


. 80978 


6 


. 10453 


. 99452 


.10510 


9.5144 


52 


.78801 


.61566 


1.2799 


.78129 


7 


.12187 


.99255 


. 12278 


8.1443 


53 


.79864 


.60182 


1.3270 


.75355 


8 


.13917 


.99027 


. 14054 


7.1154 


54 


.80902 


.58779 


1.3764 


.72654 


9 


. 15643 


. 98769 


. 15838 


6.3138 


55 


.81915 


. 57358 


1.4281 


.70021 


10 


. 17365 


.98481 


.17633 


5.6713 


56 


.82904 


.55919 


1.4826 


.67451 


11 


.19081 


.98163 


. 19438 


5.1446 


57 


.83867 


. 54464 


1 . 5399 


.64941 


12 


.20791 


.97815 


.21256 


4.7046 


58 


.84805 


. 52992 


1.6003 


. 62487 


13 


. 22495 


.97437 


.23087 


4.3315 


59 


.85717 


.51504 


1 . 6643 


. 60086 


14 


.24192 


.97030 


.24933 


4.0108 


60 


.86603 


.50000 


1.7321 


.57735 


15 


.25882 


.96593 


.26795 


3.7321 


61 


.87462 


.48481 


1.8040 


.55431 


16 


.27564 


.96126 


.28675 


3.4874 


62 


.88295 


.46947 


1.8807 


.53171 


17 


.29237 


.95630 


.30573 


3.2709 


63 


.89101 


.45399 


1 . 9626 


. 50953 


18 


. 30902 


.95106 


.32492 


3.0777 


64 


.89879 


. 43837 


2 . 0503 


.48773 


19 


.32557 


.94552 


.34433 


2 . 9042 


65 


.90631 


. 42262 


2.1445 


.46631 


20 


.34202 


. 93969 


.36397 


2.7475 


66 


.91355 


.40674 


2 . 2460 


.44523 


21 


.35837 


.93358 


.38386 


2.6051 


67 


.92050 


39073 


2 . 3559 


.42447 


22 


.37461 


.92718 


.40403 


2.4751 


68 


.92718 


.37461 


2.4751 


. 40403 


23 


.39073 


.92050 


.42447 


2.3559 


69 


.93358 


.35837 


2.6051 


. 38386 


24 


.40674 


.91355 


.44523 


2 . 2460 


70 


.93969 


. 34202 


2.7475 


.36397 


25 


.42262 


.90631 


.46631 


2 . 1445 


71 


.94552 


.32557 


2 . 9042 


.34433 


26 


.43837 


.89879 


.48773 


2.0503 


72 


.95106 


. 30902 


3.0777 


.32492 


27 


.45399 


.89101 


.50953 


1.9626 


73 


.95630 


.29237 


3.2709 


.30573 


28 


.46947 


. 88295 


.53171 


1.8807 


74 


.96126 


.27564 


3.4874 


.28675 


29 


.48481 


.87462 


.55431 


1 . 8040 


75 


.96593 


.25882 


3.7321 


.26795 


30 


.50000 


.86603 


.57735 


1.7321 


76 


.97030 


.24192 


4.0108 


.24933 


31 


.51504 


.85717 


.60086 


1.6643 


77 


.97437 


. 22495 


4.3315 


. 23087 


32 


. 52992 


. 84805 


.62487 


1.6003 


78 


.97815 


.20791 


4.7046 


.21256 


33 


. 54464 


.83867 


. 64941 


1 . 5399 


79 


.98163 


.19081 


5.1446 


: 19438 


34 


.55919 


. 82904 


.67451 


1 . 4826 


80 


.98481 


. 17365 


5.6713 


.17633 


35 


.57358 


.81915 


.70021 


1.4281 


81 


.98769 


.15643 


6.3138 


. 15838 


36 


. 58779 


. 80902 


. 72654 


1.3764 


82 


.99027 


.13917 


7.1154 


. 14054 


37 


.60182 


.79864 


. 75355 


1 . 3270 


83 


.99255 


.12187 


8.1443 


.12278 


38 


.61566 


.78801 


.78129 


1 . 2799 


84 


.99452 


. 10453 


9.5144 


.10510 


39 


.62932 


.77715 


.80978 


1.2349 


85 


.99619 


.08716 


11.430 


.08749 


40 


.64279 


. 76604 


.83910 


1.1918 


86 


.99756 


.06976 


14.301 


. 06993 


41 


.65606 


.75471 


. 86929 


1 . 1504 


87 


.99863 


.05234 


19.081 


.05241 


42 


.66913 


.74314 


. 90040 


1.1106 


88 


.99939 


. 03490 


28 . 636 


.03492 


43 


. 68200 


.73135 


. 93252 


1 . 0724 


89 


.99985 


.01745 


57.290 


.01746 


44 


. 69466 


.71934 


. 96569 


1 . 0355 


90 


1.00000 


. 00000 


00 


.00000 


45 


.70711 


.70711 


1 . 0000 


1 . 0000 













*For the purposes of our study in Practical Mathematics, the above table of values for sine, 
cosine, tangent, and cotangent of angles to degrees only is sufficient. Tables giving values to 
minutes and seconds are used in more accurate work. 



PRACTICAL MATHEMATICS 



171 



LOGARITHMS 








1 


2 


3 


4 


5 
02 1 2 


6 


7 


8 


9 


1 2 


3 


4 5 


6789 


IO 


0000 


0043 


0086 


0128 


0170 


0253 0294 033410374 


4 8 12 r 


25 29 33 37 


II 


0414 


0453 


0492 


0531 


0569 


0607 


0645 0682 0719 0755 


4 8 11 


15 19 


23 26 30 34 


12 


0792 


0828 


0864 


0899 


0934 


n\M\\) 


1004'1038 1072 


1106 


3 7 10 14 17 


21 24 28 31 


13 


1139 


1173 


1206 


1239 


1271 


1303 


1335 1367 1399 


1430 


3 6 10 13 16 


19 23 26 29 


14 


14G1 


1492 


1523 1553 


1584 


1614 


1644 1673 1703 


1732 


3 6 


9 12 15 


18 21 24 27 


15 


1761 


1790 


L818 


1847 


1875 


1903 


1931:1959 1987 


2014 


3 6 


8 11 14 


17 20 22 25 


16 


2041 


2068 


2095 


2122 


2148 


2175 


22012227 2253 


2279 


3 5 


8 11 13 


16 18 21 24 


17 


2304 


2330 2355 


23S0 


2405 


2430 


2 155 2480 2504 


2529 


2 5 


7 10 12 


15 17 20 22 


18 


2553 


25772601 


2625 


2648 


2672 


2695;2718 2742 


2765 


2 5 


7 


9 12 


14 16 19 21 


19 


27SS 


2810 2833 


2S56 


2878 


2900 
3118 


2923 

3139 


2945 2967 


2989 


2 4 


7 


9 11 


13 16 18 20 


20 


3010 


3032 


3951 


3075 


3096 


3160 3181 


3201 


2 4 


6 


8 11 


13 15 17 19 


21 


3222 


3243 


3263 


3284 


3304 


3324 


3345 3365 3385 


3404 


2 4 


6 


8 10 


12 14 16 18 


22 


3424 


3444 


3464 


3483 


3502 


3522 


354135603579 


3598 


2 4 





8 10 


12 14 15 17 


23 


3017 


3636 


3655 


3674 


3692 


3711 


3729 3747 3766 


3784 


2 4 


6 


7 9 


11 13 15 17 


24 


3802 


3S20 


3S38 


3S56 


3874 


3892 


3909 3927 


3945 


3962 


2 4 


5 


7 9 


11 12 14 16 


25 


3979 


3997 


4014 


4031 


4048 


4065 


40S2 


4099 


4116 


4133 


2 3 


5 


7 9 


10 12 14 15 


26 


4150 


4166 


41S3 


4200 


4216 


4232 


4249 


4265 


4281 


4298 


2 3 


5 


7 8 


10 11 13 15 


27 


4314 


4330 


4346 


4362 


4378 


4393 


4409 


4425 4440 


4456 


2 3 


5 


6 8 


9 11 13 14 


28 


4472 


4487 


4502 


4518 


4533 


4548 


4564'4579'4594 


4609 


2 3 


5 


6 8 


9 11 12 14 


29 


4624 


4639 


4654 


4669 


46S3 


469S 


471347284742 


4757 


1 3 


4 


6 7 


9 10 12 13 


30 


1771 


4 786 


4800 


4814 


4829 


4843 


4857 48714886 


4900 


1 3 


4 


6 7 


9 10 11 13 


31 


4914 


492S 


4942 


4955 


4969 


4983 


4997|5011 5024 


5038 


1 3 


4 


6 7 


8 10 11 12 


32 


5051 


5065 


5079 


5092 


5105 


5119 


5132 5145 5159 


5172 


1 3 


4 


5 7 


8 9 11 12 


33 


5155 


5198 


5211 


5224 


5237 


5250 


526352765289 


5302 


1 3 


4 


5 6 


8 9 10 12 


34 


5315 


5328 


5340 


5353 


5366 


5378 


5391 5403|5416 


5428 


1 3 


4 


5 6 


8 9 10 11 


35 


5441 


5453 


5465 


5478 


5490 


5502 


5514 5527 


5539 


5551 


1 2 


4 


5 6 


7 9 10 11 


36 


5563 


5575 


5587 


5599 


5611 


5623 


5635i5647 


5658 


5670 


1 2 


4 


5 6 


7 8 10 11 


37 


5682 


5694 


5705 


5717 


5729 


5740 


57525763 


5775 


5786 


1 2 


3 


5 6 


7 8 9 10 


38 


5798 


5809 


5821 


5832 


5843 


5855 


5866 5877 


5888 


5899 


1 2 


3 


5 6 


7 8 9 10 


39 


5911 


5922 5933 


5944 


5955 


5966 


59775988 


5999 


6010 


1 2 


3 


4 5 


7 8 9 10 


40 


6021 


6031.6042 


6053 


6064 


6075 


6085,6096,6107 


6117 


1 2 


3 


4 5 


6 8 9 10 


41 


6128 


61386149 


6160 


6170 


6180 


619162016212 


6222 


1 2 


3 


4 5 


6 7 8 9 


42 


6232 


6243 6253 


6263 


6274 


6284 


629463046314 


6325 


1 2 


3 


4 5 


6 7 8 9 


43 


6335 


63456355 


6365 


6375 


6385 


639564056415 


6425 


1 2 


3 


4 5 


6 7 8 9 


44 


6435 


6444 6454 6464 


6474 


6484 


6493 6503,6513 


(5522 
6618 


1 2 


3 


4 5 


6 7 8 9 


45 


6532 


6542 


6551J6561 


6571 


6580 


6590 6599,6609 


1 2 


3 


4 5 


6 7 8 9 


46 


6628 


6637 


6646 


6656 


6665 


6675 


668466936702 


6712 


1 2 


3 


4 5 


6 7 7 8 


47 


6721 


6730 


6,739 


6749 


6758 


6767 


677667856794 


6803 


1 2 


3 


4 5 


5 6 7 8 


48 


6812 


6S21 


6830 


6839 


6848 


6857 


6866 68756884 


6893 


1 2 


3 


4 4 


5 6 7 8 


49 
50 


6990 


6911 
6998 


6920 
7007 


6928 


6937 


6946 
7033 


6955'6964 6972 


6981 


1 2 


3 


4 4 


5 6 7 8 


7016 


7024 


7042 7050 7059 


7067 


1 2 


3 


3 4 


5 6 7 8 


51 


7070 


7084 


7093 


7101 


7110 


7118 


7126;7135 7143 


7152 


1 2 


3 


3 4 


5 6 7 8 


52 


7160 


7168 


7177 


7185 


7193 


7202 


7210|7218 7226 


7235 


1 2 


2 


3 4 


5 6 7 7 


53 


7243 


7251 


7259 


7267 


7275 


7284 


7292 7300 7308 


7316 


1 2 


2 


3 4 


5 6 6 7 


54 


7324 


7332 


7340 


7348 


7356 


7364 


73727380 7388 


7396 


12 


2 


3 4 


5 6 6 7 



172 



PRACTICAL MATHEMATICS 



LOGARITHMS 








1 

7412 
7490 
7566 
7642 
7716 
7789 
7860 
7931 
8000 
8069 
8136 
8202 
8267 
8331 
8395 


2 


3 


4 

7435 
7513 

7589 
7664 
7738 
7810 
7882 
7952 
8021 
8089 
8156 
8222 
8287 
8351 
8414 


5 

7443 
7520 

7597 
7672 
7745 
7818 
7889 
7959 
8028 
8096 
8162 
8228 
8293 
8357 
8420 


6 

7451 

7528 
7604 
7679 
7752 
7825 
7896 
7966 
8035 
8102 
8169 
8235 
8299 
8363 
8426 
8488 
8549 
8609 
8669 
8727 


7 

7459 
7536 
7612 
7686 
7760 
7832 
7903 
7973 
8041 
8109 
8176 
8241 
8306 
8370 
8432 
8494 
8555 
8615 
8675 
8733 


8 

7466 
7543 
7619 
7694 
7767 
7839 
7910 
7980 
8048 
8116 
8182 
8248 
8312 
8376 
8439 
8500 
8561 
8621 
8681 
8739 


9 

7474 
7551 
7627 
7701 
7774 
7846 
7917 
7987 
8055 
8122 
8189 
8254 
8319 
8382 
8445 
8506 
8567 
8627 
8686 
8745 
8802 
8859 
8915 
8971 
9025 
9079 
9133 
9186 
9238 
9289 
9340 
9390 
9440 
9489 
9538 


12 3 4 5 


6789 


55 
56 
57 
58 
59 
60 
61 
62 
63 
64 

65 
66 

67 
68 
69 
70 
71 
72 
73 
74 
75 
76 
77 
78 
79 
80 
81 
82 
83 
84 
85 
86 

87 
88 
89 
90 
9i 
92 

93 
94 
95 
96 
97 
98 
99 


7404 
74S2 
7559 
7634 
7709 
7782 
7853 
7924 
7993 
8062 
8129 
8195 
8261 
8325 
8388 


7419 7427 
7497| 7505 
7574 7582 
7649,7657 
7723,7731 
7796 7803 
7868,7875 
79387945 
8007 8014 
80758082 


12 2 3 4 
12 2 3 4 
12 2 3 4 
112 3 4 
112 3 4 


5 5 6 7 
5 5 6 7 
5 5 6 7 
4 5 6 7 
4 5 6 7 


112 3 4 
112 3 4 
112 3 3 
112 3 3 
112 3 3 


4 5 6 6 
4 5 6 6 
4 5 6 6 
4 5 5 6 
4 5 5 6 


8142 
8209 
8274 
8338 
8401 


8149 
8215 
8280 
8344 
8407 


•112 3 3 
112 3 3 
112 3 3 
112 3 3 
112 2 3 


4 5 5 6 
4 5 5 6 
4 5 5 6 

4 4 5 6 
4 4 5 6 


8451 
8513 
8573 
8633 
8692 
8751 
8808 
8865 
8921 
8976 
9031 
9085 
9138 
9191 
9243 
9294 
9345 
9395 
9445 
9494 
9542 
9590 
9638 
9685 
9731 
9777 
9823 
9868 
9912 
9956 


8457 
8519 
8579 
8639 
8698 
8756 
8814 
8871 
8927 
8982 


8463 
8525 
8585 
8645 
8704 
8762 
8820 
8876 
8932 
8987 


8470 
8531 
8591 
8651 
8710 
8768 
8825 
8882 
8938 
8993 


8476 

8537 
8597 
8657 
8716 
8774 
8831 
8887 
8943 
8998 


8482 
8543 
8603 
8663 

8722 
8779 
8837 
8893 
8949 
9004 


112 2 3 
112 2 3 
112 2 3 
112 2 3 
112 2 3 


4 4 5 6 
4 4 5 5 
4 4 5 5 
4 4 5 5 
4 4 5 5 


8785 
8842 
8899 
8954 
9009 
9063 
9117 
9170 
9222 
9274 
9325 
9375 
9425 
9474 
9523 
9571 
9619 
9666 
9713 
9759 
9805 
9850 
9894 
9939 
9983 


8791 
8848 
8904 
8960 
9015 
9069 
9122 
9175 
9227 
9279 
9330 
9380 
9430 
9479 
9528 
9576 
9624 
9671 
9717 
9763 
9809 
9854 
9899 
9943 
9987 


8797 
8854 
8910 
8965 
9020 
9074 
9128 
9180 
9232 
9284 
9335 
9385 
9435 
9484 
9533 


112 2 3 
112 2 3 
112 2 3 
112 2 3 

112 2 3 


3 4 5 5 
3 4 5 5 
3 4 4 5 
3 4 4 5 
3 4 4 5 


9036 
9090 
9143 
9196 
9248 
9299 
9350 
9400 
9450 
9499 
9547 
9595 
9643 
9689 
9736 
9782 
9827 
9872 
9917 
9961 


9042 
9096 
9149 
9201 
9253 


9047 
9101 
9154 
9206 

9258 


9053 
9106 
9159 
9212 
9263 
9315 
9365 
9415 
9465 
9513 
9562 
9609 
9657 
9703 
9750 
9795 
9841 
9886 
9930 
9974 


9058 
9112 
9165 
9217 
9269 
9320 
9370 
9420 
9469 
9518 
9566 
9614 
9661 
9708 
9754 
9800 
9845 
9890 
9934 
9978 


112 2 3 
112 2 3 
112 2 3 
112 2 3 
112 2 3 


3 4 4 5 
3 4 4 5 
3 4 4 5 
3 4 4 5 
3 4 4 5 


9304,9309 
93559360 
9405,9410 
9455'9460 
95049509 
95529557 
9600 9605 
9647J9652 
9694'9699 
974l!9745 


112 2 3 
112 2 3 
112 2 
112 2 
112 2 


3 4 4 5 
3 4 4 5 
-3 3 4 4 
3 3 4 4 
3 3 4 4 


9581 
9628 
9675 
9722 
9768 
9814 
9859 
9903 
9948 
9991 


9586 
9633 
9680 
9727 
9773 
9818 
9863 
9908 
9952 
9996 


112 2 
112 2 
112 2 
112 2 
112 2 


3 3 4 4 
3 3 4 4 
3 3 4 4 
3 3 4 4 
3 3 4 4 


9786 9791 
9832 9836 
9877,9881 
99219926 
9965 9969 


112 2 
112 2 
112 2 
112 2 
112 2 


3 3 4 4 
3 3 4 4 
3 3 4 4 
3 3 4 4 
3 3 3 4 



LIBRARY OF 



CONGRESS 




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